Category: Class 11

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

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Samacheer Kalvi 11th Chemistry Chapter 9 Solutions Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Solutions Multiple Choice Questions

Question 1.
The molality of a solution containing 1 .8g of glucose dissolved in 250g of water is …………
(a) 0.2 M
(b) 0.01 M
(c) 0.02 M
(d) 0.04 M
Answer:
(d) 0.04 M
Solution:

Question 2.
Which of the following concentration terms is/are independent of temperature?
(a) molality
(b) molarity
(c) mole fraction
(d) (a) and (c)
Answer:
(d) (a) and (c)
Solution:
Molality and mole fraction are independent of temperature.

Question 3.
Stomach acid, a dilute solution of HCI can be neutralised by reaction with Aluminium hydroxide
Al(OH)₃ + 3HCl(aq) → AlCl₃ + 3H₂O
How many millilitres of 0.1 M Al(OH)₃ solution are needed to neutralise 21 mL of 0.1 M HCl
(a) 14 mL
(b) 7 mL
(c) 21 mL
(d) none of these
Answer:
(b) 7 mL
Solution:
M₁ x V₁ = M₂ x V₂
∵ 0.1 M Al(OH)₃ gives 3 x 0.1 = 0.3 M OH^– ions .
0.3 x V₁ = 0.1 x 21
V₁= \(\frac { 0.1 x 21 }{ 0.3 }\) = 7ml

Question 4.
The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 x 10⁴ atm at 300K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300K?
(a) 1 x 10^-4
(b) 1 x 10^-6
(c) 2 x 10^-5
(d) 1 x 10^-5
Answer:
(d) 1 x 10^-5
Solution:
P_N2 = 0.76atm
K_H = 7.6 x 10⁴
x = ?
P_N2 = K_H . x
0.76 = 7.6 x 10⁴x x
x = \(\frac { 0.76 }{ 7.6\times { 10 }^{ 4 } }\) = 1 x 10^-5

Question 5.
The Henry’s law constant for the solubility of Nitrogen gas in water at 350K is 8 x 10⁴ atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350K and 4 atm pressure is ………….
(a) 4 x 10^-4
(b) 4 x 10⁴
(c) 2 x 10^-2
(d) 2.5 x 10^-4
Answer:
(d) 2.5 x 10^-4
Solution:
K_H = 8 x 10⁴
(x_N2 )_{in air} = 0.5
Total pressure = 4 atm
Partial pressure of nitrogen = Mole fraction Total pressure
= O.5 x 4 = 2

Question 6.
Which one of the following is incorrect for ideal solution?
(a) ∆H_mix = 0
(b) ∆U_mix = o
(c) ∆P = P_Observed – P_{Calculated by raoults law} = 0
(d) ∆G_mix = 0
Answer:
(d) ∆G_mix = 0
Solution:
For an ideal solution, ∆S_mix \(\neq\) 0; Hence ∆G_mix \(\neq\) 0
∴ Incorrect is ∆G_mix = 0

Question 7.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N₂
(b) He
(c) CO₂
(d) H₂
Answer:
(c) CO₂
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.

Question 8.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………
(a) P₁ + x₁(P₂ – P₁)
(b) P₂ – x₁(P₂ + P₁)
(c) P₁ – x₂(P₁ – P₂)
(d) P₁ + x₂(P₁ – P₂)
Answer:
(c) P₁ – x₂(P₁ – P₂)
Solution:
P_total = P₁ + P₂
= P₁ x₁ + P₂x₂
= P₁(1 – x₂) + P₂x₂
= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂(P₁ – P₂)
[∵x₁ + x₂ = 1
x₁ = 1 – x₂]

Question 9.
Osomotic pressure (π) of a solution is given by the relation ……………
(a) π = nRT
(b) πV = nRT
(c) πRT = n
(d) none of these
Answer:
(b) πV = nRT
Solution:
n = CRT
n = \(\frac { n }{ V }\)
π V = nRT

Question 10.
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?
(a) Acetone + chloroform
(b) Water + nitric acid
(c) HCI + water
(d) ethanol + water
Answer:
(d) ethanol + water

Question 11.
The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B is 0.2. The ratio of mole fraction of B and A dissolved in water will be …………
(a) \(\frac { 2x }{ y }\)
(b) \(\frac { y }{ 0.2x }\)
(c) \(\frac { 0.2x }{ y }\)
(d) \(\frac { 5x }{ y }\)
Answer:
(d) \(\frac { 5x }{ y }\)
Solution:
Given,

Question 12.
At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732mm. If K_b = 0.52, the boiling point of this solution will be …………..
(a) 102°C
(b) 100°C
(c) 101°C
(d) 100.52°C
Answer:
(c) 101°C
Solution:
\(\frac { ΔP }{ P° }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } }\)
W₂ = 6.5g
W₁ = 100g
K_b = 0.52

Question 13.
According to Raoults law, the relative Lowering of vapour pressure for a solution is equal to….
(a) molefraction of solvent
(b) mole fraction of solute
(c) number of moles of solute
(d) number of moles of solvent
Answer:
(b) mole fraction of solute
Solution:
\(\frac { ∆P }{ P° }\) = x₂ (Mole fraction of the solute)

Question 14.
At same temperature. which pair of the following solutions are isotonic?
(a) 0.2 M BaCl₂ and 0.2M urea
(b) 0.1 M glucose and 0.2 M urea
(c) 0.1 MNaCl and 0.1 MK₂SO₄
(d) 0.1 MBa(NO₃)₂ and 0.1 MNa₂ SO₄
Answer:
(d) 0.1 M Ba (NO₃)₂ and 0.1 M Na₂ SO₄
Solution:
0.1 x 3 ion [Ba² + 2NO₃^–], 0.1 x 3 ion [2Na⁺, SO₄^–]

Question 15.
The empirical formula of a non-electrolyte(X) is CH₂O. A solution containing six gram of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is
(a) C₂H₄O₂
(b) C₈H₁₆O₈
(c) C₄H₈O₄
(d) CH₂O
Answer:
(b) C₈H₁₆O₈
Solution:
(π₁)_{non electrolute} = (π₂)_glucose
C₁RT = C₂RT

\(\frac { 6 }{ n(30) }\) = 0.025
n = \(\frac { 6 }{ 0.025 x 30 }\) = 30
∴ Molecular formula C₈H₁₆O₈

Question 16.
The K_H for the solution of oxygen dissolved in water is 4 x 10⁴ atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is …………..
(a) 4.6 x 10³
(b) 1.6 x 10⁴
(c) 1 x 10^-5
(d) 1 x 10⁵
Answer:
(c) 1 x 10^-5
Solution:
K_H = 4 x 10⁴ atm,
(P_O2)_air = 0.4 atm,
(x_o2)_{in solution} = ?
air – in solution
(P_O2)_air = K_H(x_o2)_{in solution}
0.4 = 4 x 10⁴(x_o2)_{in solution}
(x_o2)_{in solution} = \(\frac { 0.4 }{ 4\times { 10 }^{ 4 } }\) = 1 x 10^-5

Question 17.
Normality of 1.25M sulphuric acid is …………
(a) 1.25 N
(b) 3.75 N
(c) 2.5 N
(d) 2.25 N
Answer:
(c) 2.5 N
Solution:
Normality of H₂SO₄ = (No. of replacable H⁺) x M = 2 x 1.25 = 2.5 N

Question 18.
Two liquids X and Y on mixing gives a warm solution. The solution is …………..
(a) ideal
(b) non-ideal and shows positive deviation from Raoults law
(c) ideal and shows negative deviation from Raoults Law
(d) non – ideal and shows negative deviation from Raoults Law
Answer:
(d) non – ideal and shows negative deviation from Raoults Law
Solution:
∆H_mix is negative and show negative deviation from Raoults law.

Question 19.
The relative lowering of vapour pressure of a sugar solution in water is 3.5 x 10^-3. The mole fraction of water in that solution is …………
(a) 0.0035
(b) 0.35
(c) 0.0035/18
(d) 0.9965
Answer:
(d) 0.9965
Solution:

Question 20.
The mass of a non-volatile solute (molar mass 80 g mol^-1) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90% ………..
(a) 10g
(b) 20g
(c) 9.2 g
(d) 8.89g
Answer:
(d) 8.89g
Solution:

Question 21.
For a solution, the plot of osmotic pressure (π) verses the concentration (e in mol L^-1) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is ………..
(a) 310 x 0.082 K
(b) 3 10°C
(c) 37°C
(d) \(\frac { 310 }{ 20.082}\)
Answer:
(c) 37°C
Solution:
π = CRT
y = x(m)
m = RT
310 R = RT
T = 310 K
= 37°C

Question 22.
200 ml of an aqueous solution of a protein contains 1 .26g of protein. At 300K, the osmotic pressure of this solution is found to be 2.52 x 10^-3 bar. The molar mass of protein will be (R =0.083 Lhar mol^-1 K^-1) ……………
(a) 62.22 Kg mol^-1
(b) 12444 g mol^-1
(c) 300g mol^-1
(d) none of these
Answer:
(a) 62.22 Kg mol^-1
Solution:
π = CRT
M = \(\frac { WRT }{ π1 }\) = \(\frac { 1.26\times 0.083\times 300 }{ 2.52\times { 10 }^{ -3 }\times 0.2 }\) = 62.22Kg mol^-1

Question 23.
The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is ………..
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Solution:
Ba(OH)₂ dissociates to form Ba²⁺ and 2OH^-1 ion
α = \(\frac { (i – 1) }{ (n – 1) }\)
i = α (n – 1) + 1
n = i = 3 ( for Ba (OH)₂, α = 1 )

Question 24.
What is the molality of a 10% w/w aqueous sodium hydroxide solution?
(a) 2.778
(b) 2.5
(c) 10
(a) 0.4
Answer:
(b) 2.5
Solution:
100% \(\frac { w }{ w }\) aqueous NaOH solution means that 10 g of sodium hydroxide in 100g solution.

Question 25.
The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is ………
(a) α = \(\frac { n(i – 1) }{ n – 1 }\)
(b) α² = \(\frac { n(1 – i) }{ n – 1 }\)
(c) α = \(\frac { n(i – 1) }{ 1 – n }\)
(d) α = \(\frac { n(1 – i) }{ n(1 – i) }\)
Answer:
(c) α = \(\frac { n(i – 1) }{ 1 – n }\)
Solution:
α = \(\frac { (i – 1)n }{ (n – 1) }\) (or) \(\frac { n(i – 1) }{ (1 – n) }\)

Question 26.
Which of the following aqueous solutions has the highest boiling point?
(a) 0.1 M KNO₃
(b) 0.1 M Na₃PO₄
(c) 0.1 M BaCl₂
(d) 0.1 M K₂SO₄
Answer:
(a) 0.1 M KNO₃
Solution:
Elevation of boiling point is more in the case of Na₃PO₄(no. of ions 4; 3 Na⁺, PO₄^3-)

Question 27.
The freezing point depression constant for water is 1.86° k kg mo1^-1 . If 5g Na₂SO₄ is dissolved in 45g water, the depression in freezing point is 3.64°C. The van’t Hoff factor for Na₂SO₄ is ……..
(a) 2.50
(b) 2.63
(c) 3.64
(d) 5.50
Answer:
(a) 2.50
Solution:
K_f = 1.86
W₂ = 5g
∆T_f = 3.64
M₂ = 142
W₁ = 45g
ΔT_f = i x K_f

Question 28.
Equimolal aqueous solutions of NaCI and KCI are prepared. If the freezing point of NaCI is – 2°C, the freezing point of KCI solution is expected to be ………
(a) – 2°C
(b) – 4°C
(c) – 1°C
(d) 0°C
Answer:
(a) – 2°C
(b) – 4°C
(c) – 1°C
(d) 0°C
Solution:
Equimolal aqueous solution of KCI also shows 2° C depression in freezing point.

Question 29.
Phenol dimerises in henzene having van’t Hoff factor 0.54. What is the degree of association?
(a) 0.46
(b) 92
(c) 46
(d) 0.92
Answer:
(d) 0.92
Solution:
α = \(\frac { (1-i)n }{ (n-1) }\) = \(\frac { (1 – 0.54)2 }{ (2 – 1) }\) = 0.46 x 2 = 0.92
Question 30.
Assertion : An ideal solution obeys Raoults Law
Reason : In an ideal solution, solvent-solvent as well as solute-solute interactions are similar to solute-solvent interactions.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) both assertion and reason are true and reason is the correct explanation of assertion

Samacheer Kalvi 11th Chemistry Solutions Short Answer Questions

Question 31.
Define

1. Molality
2. Normality

Answer:
1. Molality (m):
It is defined as the number of moles of the solute present in 1 kg of the solvent

2. Normality (N):
It is defined as the number of gram equivalents of solute in I litre of the solution.

Question 32.
What is a vapour pressure of liquid? What is relative lowering of vapour pressure?
Answer:
1. The pressure of the vapour in equilibrium with its liquid ¡s called vapour pressure of the liquid at the given temperature.

2. The relative lowering of vapour pressure is defined as the ratio of lowering of vapour. pressure to vapour pressure of pure solvent. Relative lowering of vapour pressure

Question 33.
State and explain Henry’s law.
Answer:
Henry’s law:
This law states “that the partial pressure of the gas in vapour phase is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations.”
P_solute ∝ x_{solute in solution}
P_solute = K_H. x_{solute in solution}
x_solute = mole fraction of solute in the solution
K_H = empirical constant.
P_solute = Vapour pressure of the solute (or) the partial pressure of the gas in vapour state. The value of K_H depends on the nature of the gaseous solute and solvent.

Question 34.
State Raoult law and obtain expression for lowering of apour pressure when nonvolatile solute is dissolved in solvent.
Answer:
Raoult’s law:
This law states that “in the case of a solution of volatile liquids the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction.
P_A ∝ x _A
when x_A = 1,
then k = P°_A
(P°A = vapour pressure of pure component)
P_A = P°_A . xa
P_B = P°_B . xb
when a non volatile is dissolved in pure water, the vapour pressure of the pure solvent will decrease. In such solution, the vapours pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.
P_solution ∝ x_A
P_solution = k . x_A
x_A = 1, k = P°_solvent
P°_solution = P°_solvent – P_solution
Lowering of vapour pressure = P°_solvent – P_solution
Relative lowering of vapour pressure = \(\frac { P° – P }{ P° }\) = x_B
where x_B = Mole fraction of solute.

Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
K_f = molar freezing point depression constant or cryoscopic constant.
∆T_f = K_f . m,
where
∆T_f = depression in freezing point.
m = molality of the solution
K_f = cryoscopic constant
If m = I
∆T_f = K_f
i.e., cryoscopic constant is equal to the depression in freezing point for 1 molal solution cryoscopic constant depends on the molar concentration of the solute particles. K_f is directly proportional to the molal concentration of the solute particles.

W_B = mass of the solute
W_A = mass of solvent
M_B = molecular mass of the solute.

Question 36.
What is osmosis?
Answer:
Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.

Question 37.
Define the term bisotonic
Answer:
1. Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

2. When such solutions arc separated by a semipermeable membrane, solvent flow between one to the other on either direction is same. i.e.. the net solvent flow between two isotonic solutions is zero.

Samacheer Kalvi 11th Chemistry Solutions Long Answer Questions

Question 38.
You are provided with a solid ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
1. Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

2. Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

3. Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCI in 1 litre of watcr at 25°C.
2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCI in 1 litre of water at 25°C.
3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCI in 1 litre of water at 25°C.

Question 39.
Explain the effect of pressure on the solubility.
Answer:
1. The change in pressure does not have any significant effect in the solubility of solids and liquids as they are not compressible. However the solubility of gases generally increases with increase of pressure.

2. According to Le – chatlier’s principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more number of gaseous molecules dissolves in the solvent.

3. If pressure increases, solubility of gas also increases.

Question 40.
A sample of 12 M Concentrated hydrochloric acid has a density 1.2 gL^-1. Calculate the molality.
Answer:
Given:
Molarity = 12 M HCI
Density of the solution = 1.2 g L^-1
In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.

Calculate mass of water (solvent)
Mass of 1 litre HCI solution = density x volume
= 1.2gmL^-1 x 1000 mL = 1200g
Mass of LICI = No. of moles of HCI x molar mass of HCI
= 12mol x 36.5 g mol^-1 = 438g
Mass of waler = mass of HCI solution – mass of HCI
Mass of waler = 1200 – 438 = 762 g
Molalily =\(\frac { 12 }{ 0.762 }\) = 15.75m

Question 41.
A 0.25 M glucose solution at 370.28 K has approximately the pressure as blood. What is the osmotic pressure of blood?
Solution.
C = 0.25 M
T = 37O.28 K
(π)_gIucose = CRT
(π) = 0.25 mol L^-1 x 0.082 L atm K^-1 morl^-1 x 370.28K
= 7.59 atm

Question 42.
Calculate the molality of a solution containing 7.5g of glycine (NH₂ – CH₂ – COOH) dissolved in 500g of water.
Solution:

Question 43.
Which solution has the lower freezing point? 10g of methanol (CH₃OH) in 100g of water (or) 20g of ethanol (C₂H₅HO) In 200g of water.
Solution:
∆T_f = K_f . m i.e. ∆T_f ∝ m

∴ Depression in freezing point is more in methanol solution and it will have lower freezing point.

Question 44.
How many moles of solute particles are present in one litre of 10^-4 M potassium sulphate?
Solution:
In 10^-4M K₂SO₄ solution, there are 10^-4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2 K⁺ and 1SO₄^2-)
1 mole of K₂SO₄ contains 3 x 6.023 x 10²³ ions
10 mole of K₂SO₄ contains 3 x 6.023 x 10² x 10^-4 ions = 18.069 x 10¹⁹

Question 45.
Henry’s law constant for solubility of methane in benzene is 4.2 x 10^-5 mm Hg at a particular constant temperature. At this temperature, calculate the solubiiitv of methane at

1. 75O mm Hg
2. 84OnimHg

Solution:
(K_H)_Benzene = 4.2 x 10^-5 mm Hg. Solubility of methane = ? P = 750 mm Hg, p = 840 mm Hg
According to Henry’s Law,
P = K_H . x_solution
750 mm Hg = 4.2 x 10^-5 mm Hg . x_solution

Question 46.
The observed depression in freezing point of water for a particular solution is 0.093°C. Calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K kg mol^-1.
Solution:
T₁= 0.093°C = 0.093K
m = ?
K_f = 1.86K kg mol^-1
∆T_f = k_f . m

Question 47.
The vapour pressure of pure henzene (C₆H₆) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution:
P⁰_C6H6 = 640 mm Hg
W₂ = 2.2 g (non volatile solute)
W₁ = 40 g (benzene)
P_solution = 600 mm Hg
M₂ = ?

Samacheer Kalvi 11th Chemistry Solutions In Text Questions – Evaluate Yourself

Question 1.
If 5.6 g of KOH is present in (a) 500 mL and (b) I litre of solution, calculate the molarity of each of these solutions.
Solution.
Mass of KOH = 5.6g
No. of moles = \(\frac { 5.6 }{ 5.6 }\) = 0.1 mol
1. Volume of the solution = 500 ml = 0.5 L

2. Volume of the solution = IL

3. Volume of the solution = IL
Molarity = \(\frac { 0.1 }{ 1 }\) M

Question 2.
2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water.
Solution:
Mass of glucose = 2.82 g
No. of moles of glucose = \(\frac { 2.82 }{ 180 }\) = 0.0 16
Mass of water = 30g = \(\frac { 30 }{ 18 }\) = 1.67
x_H2O = \(\frac { 1.67 }{ 1.67 + 0.016 }\) = \(\frac { 1.67 }{ 1.686 }\) = 0.99
x_H2O + x_glucose = 1
0.99 + x_glucose = 1
x_glucose = 1 – 0.99 = 0.01

Question 3.
The antiseptic solution of iodopovidone for the use of external application contains 10% w/v of iodopovidone. Calculate the amount of iodopovidone present in a typical dose of 1.5 mL.
Solution:
10% \(\frac { w }{ v }\) means that 10 g of solute in 100 ml solution
∴ Amount of iodopovidone in 1.5 ml = \(\frac { 10g }{ 100ml }\) x 1.5 ml = 0.15 g

Question 4.
A litre of sea water weighing about 1.05 kg contains 5 mg of dissohed oxygen (O₂). Express the concentration of dissolved oxygen in ppm.
Solution:

Question 5.
Describe how would you prepare the following solution from pure solute and solvent

1. 1 L of aqueous solution of 1.5 M COCI₂.
2. 500 mL of 6.0 % (v/v) aqueous methanol solution.

Solution:

1. mass of 1.5 moles of COCI₂ = 1.5 x 129.9 = 194.85g
2. 194.85g anhydrous cobalt chloride is dissolved in water and the solution is make up to one litre in a standard flask.

Question 6.
How much volume of 6 M solution of NaOH is required to prepare 500 mL of 0.250 M NaOH solution.
Solution:
6% \(\frac { v }{ v }\) aqueous solution contains 6g of methanol in 100 ml solution. To prepare 500 ml of 6% v/v solution of methanol 30g methanol is taken in a 500 ml standard flask and required quantity of water is added to make up the solution to 500 ml.

Question 7.
Calculate the proportion of O₂ and N₂ dissolved in water at 298 K. When air containing 20% O₂ and 80% N₂ by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O₂) = 4.6 x atm and K_H (N₂) 8.5 x 10⁴ atm.
Solution:
C₁V₁ = C₂V₂
6M (V₁) = 0.25M x 500 ml
V₁ = \(\frac { 0.25 x 500 }{ 6 }\)
V₁ = 20.3 mL

Question 8.
Explain why the aquatic species are more comfortable in cold water during winter season rather than warm water during the summer.
Solution:
Total pressure = 1 atm
P_N2 = \((\frac { 80 }{ 100 })\) x Total pressure = \(\frac { 80 }{ 100 }\) x 1 atm = 0.8 atm
P_O2 = \((\frac { 20 }{ 100 })\) x 1 = 0.2 atm
According to Henry’s Law
P_solute = K_H x _{solute in solution}
P_N2 = (K_H)_Nitrogen x Mole fraction of Nitrogen in solution

Question 9.
Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.
Solution:
P⁰_{pure benzene} = 50.71 mm Hg
P⁰_nepthalene = 32.06 mm Hg
Number of moles of benzene = \(\frac { 39 }{ 78 }\) = 0.5 mol
Number of moles of naphthalcne = \(\frac { 128 }{ 128 }\) =1 mol
Mole fraction of benzene = \(\frac { 0.5 }{ 1.5 }\) = 0.33
Mole fraction of naphthalene = 1 – 0.33 = 0.67
Partial vapour pressure of benzene =P⁰_benzene x Mole fraction of benzene
= 50.71 x 0.33 = 16.73 mm Hg
Partial vapour pressure of naphthalene = 32.06 x 0.67 = 21.48mm Hg
Mole fraction of benzene in vapour phase = \(\frac { 16.73 }{ 16.73 + 21.48 }\) = \(\frac { 16.73 }{ 38.21 }\) = 0.44
Mole fraction of naphthalene in vapour phase = 1 – 0.44 = 0.56

Question 10.
Vapour pressure of a pure liquid A is 10.0 torr at 27°C. The vapour pressure is lowered to 9.0 torr on dissolving one grani of B in 20g of A. If the molar mass of A is 200 g mol^-1 then calculate the molar mass of B.
Solution:
P⁰_A = 10 torr
P_solution = 9 torr
W_A = 20 g
W_B = 1 g
M_A = 200 g mol^-1
M_B = ?

Question 11.
2.56g of Sulphur is dissolved in 100g of carbon disuiphide. The solution boils at 319.692K. What is the molecular formula ofSulphur in solution? The boiling pointof CS₂ is 319. 450K. Given that Kb for CS₂ = 2.42 K kg mol^-1
Solution:
W₂ = 2.56g
W₁ = 100g
T = 319.692 K
K_b = 2.42 K kg mol^-1
∆T_b = (319.692 – 319.450) K = 0.242 K

M₂ = 256g mol^-1
Molecular mass of sulphur in solulion = 256 g mol^-1
Atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \(\frac { 256 }{ 32 }\) = 8
Hence, molecular tòrmula of sulphur is S₈.

Question 12.
2g of a non electrolyte solute dissolved in 75g of benzene lowered the freezing point of benzene by 0.20 K. The freezing point depression constant of benzene is 5.12 K Kg mol^-1. Find the molar mass of the solute.
Solution:
W₂ = 2g
W₁ = 75g
∆T_f = 0.2 K
k_f = 5.12 K kg mol^-1
M₂ = ?

Question 13.
What is the mass of glucose (C₆H₁₂O₆) in it one litre solution is isotonic with 6g L^-1 of urea (NH₂CONH₂)?
Solution:
Osmotic pressure of urea solution (π₁) = CRT
\(\frac { { W }_{ 2 } }{ { M }_{ 2 }V }\)RT = \(\frac { 6 }{ 60 x 1 }\) x RT
Osmotic pressure of glucose solution
(π₂) \(\frac { { W }_{ 2 } }{ 180\times 1 }\) x RT
For isotonic solution, π₁ = π₂
\(\frac { 6 }{ 60 }\) = \(\frac { { W }_{ 2 } }{ 180\times 1 }\) RT ⇒ W₂ = \(\frac { 6 }{ 60 }\) x 180
W₂ = 18 g

Question 14.
0.2m aqueous solution of KCI freezes at – 0.68°C calculate van’t Hoff factor. K_f for water is 1.86 K kg mol^-1.
Solution:

Given,
∆T_f = 0.680 K
m = 0.2 m,
∆T_f (observed) = 0.680K
∆T_f(Calculated) = k_f
m = 1.86 K kg mol^-1 x 0.2 mol kg^-1 = 0.372K

Samacheer Kalvi 11th Chemistry Solutions Example problems Solved

Question 1.
What volume of 4M HCI and 2M HCI should be mixed to get 500 mL of 2.SM HCI?
Solution:
Let the volume of 4M HCl required to prepare 500 mL of 2.5 M HCI = x mL
Therefore, the required volume of 2M HCI = (500 – x) mL
We know from the equation x = \(\frac { 250 }{ 2 }\) = 125 mL
Hence, volume of 4M HCI required = 125 mL
Volume of 2M HCl required = (500 – 125) mL = 375 mL

Question 2.
0.24g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature.
Solution:
P_solute = K_H x_{solute in solution}
At pressure 1.5 atm, p₁ = K_H x₁ ………..(1)
At pressure 6.0 atm, p₂ = K_Hx₂ …………..(2)
Dividing equation (1) by (2)
Weget
\(\frac { { P }_{ 1 } }{ { P }_{ 2 } }\) = \(\frac { { x }_{ 1 } }{ { x }_{ 2 } }\)
\(\frac { 1.5 }{ 6.0 }\) = \(\frac { { 0.24 } }{ { x }_{ 2 } }\)
Therefore
x₂ = \(\frac { 0.24×6.0 }{ 1.5 }\) = 0.96 g/L

Question 3.
An aqueous solution of 2% nonvolatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molar mass of the solute when P_A° is 1.013 bar?
Solution:

In a 2% solution weight of the solute is 2g and solvent is 98g
ΔP = P_A⁰ – P_solution = 1.013 – 1.004 bar = 0.009 bar

Question 4.
0.75 g of an unknown substance is dissolved in 200 g solvent. If the elevation of boiling point is 0.15 K and molal elevation constant is 7.5K kg more then, calculate the molar mass of unknown substance.
Solution:
∆T_b = K_b m = K_b x W₂ x 1000/M₂ x W₁
M₂ = K_b x W₂ x 1000/∆T_b x W₁
= 7.5 x 0.75 x 1000/0.15 x 200 = 187.5g mol^-1

Question 5.
Ethylene glycol (C₂H₆O₂) can be used as an antifreeze in the radiator of a car. Calculate the temperature when ice will begin to separate from a mixture with 20 mass percent of glycol in water used in the car radiator. K_f for water = 1.86 K kg mol^-1 and molar mass of ethylene glycol is 62g mol^-1.
Solution:
Weight of solute (W₂) = 20 mass percent of solution means 20g of ethylene glycol
Weight of solvent (water) W₁ = 100 – 20 = 80g

The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e. 7.5 K lower than the normal freezing point of water (273 – 7.5)K = 265.5K

Question 6.
At 400K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculate the molar mass of the unknown substance.
Solution:

Question 7.
The depression in freezing point is 0.24K obtained by dissolving 1g NaCI in 200g water. Calculate van’t – Hoff factor. The molal depression constant is 1.86 K kg mol^-1.
Solution:
Sol. Molar mass of solute

Samacheer Kalvi 11th Chemistry Solutions Additional Questions Solved

Samacheer Kalvi 11th Chemistry Solutions 1 Mark Questions and Answers

I. Choose the correct answer.

Question 1.
Among the following, which one is mostly present in sea water?
(a) NaCI
(b) Nal
(c) KCI
(d) MgBr₂
Answer:
(a) NaCI

Question 2.
Statement I: The most common property of sea water and air is homogeneity.
Statement II: The homogeneity implies uniform distribution of their constituents through the mixture.
(a) Statements I and II arc correct and II is the correct explanation of I.
(b) Statements I and II are correct but II is not the correct explanation of I.
(c) Statement I is correct but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) StatementI I and II are correct and II is the correct explanation I.

3. Which one of the following is a homogeneous mixture?
(a) Sea water
(b) Air
(c) Alloys
(d) All the above
Answer:
(d) All the above

Question 4.
Statement I: Salt solution is an aqueous solution.
Statement II: If water is used as the solvent, the resultant solution is called an aqueous solution.
(a) Statements I and II are correct but II is not the correct explanation of I.
(b) Statements I and II are correct and II is the correct explanation of I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(b) Statements I and II are correct and II is the correct explanation of I.

Question 5.
Statement I: The dissolution of ammonium nitrate increases steeply with increase in temperature.
Statement II: The dissolution process of ammonium nitrate is endothermic in nature.
(a) Statement I and II are correct and statement II is the correct explanation of statement I.
(b) Statement I and II are correct but II is not the correct explanation of I.
(c) Statement I is correct but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) Statement I and II are correct and statement II is the correct explanation of statement I.

Question 6.
In which of the following compound the solubility decreases with increase of temperature?
(a) sodium chloride
(b) ammonium nitrate
(c) cerie sulphate
(d) calcium chloride
Answer:
(c) ceric sulphate

Question 7.
Which of the following is not an ideal solution?
(a) Benzene & toluene
(b) n – Hexane & n – Heptane
(c) Ethyliodide & ethyl bromide
(d) Ethanol and water
Answer:
(d) Ethanol and water

Question 8.
Which one of the following shows positive deviation from Raoult’s law?
(a) Ethyliodide and Ethyl bromide
(b) Ethyl alcohol and cyclohexane
(c) Chioro benzene & bromo benzene
(d) Benzene & toluenc
Answer:
(b) Ethyl alcohol and cyclohexane

Question 9.
Which one of the following is not an non-ideal solution showing positive deviation?
(a) Benzene & acetone
(b) CCl₄ & CHCI₃
(c) Acetone & ethyl alcohol
(d) Benzene and toluene
Answer:
(d) Benzene and toluene

Question 10.
Which of the following shows negative deviation from Raoults law?
(a) Phenol and aniline
(b) Benzene and toluene
(c) Acetone and ethanol
(d) Bcnzene and acetone
Answer:
(a) Phenol and aniline

Question 11.
Which of the following is not an non-ideal solution showing negative deviation?
(a) Phenol and aniline
(b) Ethanol and water
(c) Acetone + Chlorotorm
(d) n – Heptane and n – Hexane
Answer:
(d) n – Heptanc and n – Hexane

Question 12.
Statement I: A solution of potassium chloride in water deviates from ideal behavior.
Statement II: The solute dissociates to give K and Cl ion which form strong ion dipole interaction with water molecules.
(a) Statement I & II are correct and II is the correct explanation of I
(b) Statement I & II are correct but II is not correct explanation of I
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(a) Statement I & II are correct and II is the correct explanation of I

Question 13.
Statement I: Acetic acid deviates from ideal behaviour.
Statement II: Acetic acid exists as a dimer by forming inter molecular hence deviates from Raoults law.
(a) Statement I & II are correct and II is the correct explanation of I.
(b) Statement I & II are correct but II is not the correct explanation of I.
(c) Statement I is true but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) Statement I & II are correct but II is the correct explanation of I.

Question 14.
Which one of the following has found to have abnormal molar mass? hydrogen bonds and
(a) NaCl
(b) KCI
(c) Acetic acid
(d) all the above
Answer:
(d) All the above

Question 15.
What would be the value of van’t Hoff factor for a dilute solution of K₂SO₄ in water.
(a) 3
(b) 2
(c) 1
(d) 4
Answer:
(a) 3
Solution:
ions produced = n = 3
Since
K₂SO₄ → 2K⁺ + SO₄^2-
K₂SO₄ is completely dissociated so
∝ = \(\frac { i – 1 }{ n – 1 }\) = \(\frac { i – 1 }{ 3 – 1 }\) = 1
i – 1 = 1 x 2
i – 1 = 2
i = 2+1 = 3

Question 16.
In the determination of molar mass of AB using a colligative property, what may be the value of van’t Hoff factor if the solute is 50% dissociates?
(a) 0.5
(b) 1.5
(c) 2.5
(d) 1
Answer:
(b) 1.5
Solution:
∝ = \(\frac { i – 1 }{ n – 1 }\) = 0.5
\(\frac { i – 1 }{ 2 – 1 }\) = 0.5
i – 1 = 0.5
i = 0.5 + 1 = 1.5

Question 17.
Which of the following solution has the highest boiling point?
(a) 5.85% solution of NaCI
(b) 18.0% solution of glucose
(c) 6.0% solution of urea
(d) All have same boiling point
Answer:
(a) 5.85% solution of NaCl

Question 18.
Which one of the following pair is called an ideal solution?
(a) nicotine – water
(b) water – ether
(c) water – alcohol
(d) Chiorobenzene – bromobenzene
Answer:
(d) Chiorobenzene – bromobenzene

Question 19.
Which of the following is not a colligative property?
(a) optical activity
(b) osmotic pressure
(c) elevation boiling point
(d) depression in freezing point
Answer:
(a) optical activity

Question 20.
On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?
(a) Sugar crystals in cold water
(b) Sugar crystals in hot water
(c) powdered sugar in cold water
(d) powdered sugar in hot water
Answer:
(d) powdered sugar in hot water

II. Match the following.
Question 1.

Answer:
(a) 3 4 1 2

Question 2.

Answer:
(d) 3 4 2 1

Question 3.

Answer:
(c) 2 4 1 3

Question 4.

Answer:
(a) 4 3 1 2

Question 5.

Answer:
(b) 2 4 1 3

III. Fill in the blanks.

Question 1.
……… covers more than 70% of the earth’s surface.
Answer:
Seawater

Question 2.
……… is an important naturally occurring solution.
Answer:
Air

Question 3.
An example of solid homogeneous mixture is ……….
Answer:
Brass

Question 4.
A mixture of N₂, O₂, CO₂ and other traces of gases is known as ………
Answer:
Air

Question 5
……… a non – aqueous solution.
Answer:
Br₂ in CCl₄

Question 6.
……… is an example for gaseous solution.
Answer:
Camphor in nitrogen gas

Question 7 .
……… is used for dental filling.
Answer:
Amalgam of potassium

Question 8.
Carbonated water is an example for ………
Answer:
Liquid solution

Question 9.
Humid oxygen is an example of ………
Answer:
Gaseous solution

Question 10.
The concentration of commercially available H₂O₂ is ………
Answer:
3%

Question 11.
The molality of the solution containing 45g of glucose dissolved in 2kg of water is ………
Answer:
0.125m

Question 12.
5.845 g of NaCl is dissolved in water and the solution was made up to 500 mL using a standard flask. The strength of the solution in molarity is ………
Answer:
0.2 M
Solution:

Question 13.
3.15 g of oxalic acid dihydrate is dissolved in water and the solution was made up to 100 ml using a standard flask. The strength of the solution in normality is ………
Answer:
0.5N
Solution:

Question 14.
5.85 g of NaCI is dissolved in water and the solution was made upto 500 ml using a standard flask. The strength of the solution in formality is ………
Answer:
0.2 F
Solution:

Question 15.
Neomycin, amino glycoside antibiotic cream contains 300 mg of neomycin sulphate the active ingredient in 30 g of oinment base. The mass percentage of neomycin is ………
Answer:
1%
Solution:
The mass percentage of neomycin

Question 16.
0.5 mole of ethanol is mixed with 1.5 mole of water. Then the mole fraction of ethanol and water are ……….
Answer:
0.25, 0.75
Solution:

= \(\frac { 0.5 }{ 1.5 + 0.5 }\) = \(\frac { 0.5 }{ 2.0 }\) = 0.25
Mole fraction of water = \(\frac { 1.5 }{ 2.0 }\) = 0.75

Question 17.
50 mL of tincture of benzoin, an antiseptic solution contains 10 ml of benzoin. The volume percentage of benzoin is ……….
Answer:
20%
Solution:
Volume percentage of benzoin

= \(\frac { 10 }{ 50 }\) x 100 = 20%

Question 18.
A 60 ml of paracetamol pediatric oral suspension contains 3g of paracetamol. The mass percentage of paracciamol is …………
Answer:
5%
Solution:
Mass percentage of paracetamol =

= \(\frac { 3 }{ 60 }\) x 100 = 5%

Question 19.
50 ml of tap water contains 20 mg of dissolved solids. The TDS value in ppm is ………..
Answer:
400 ppm
Solution:

Question 20.
The concentration term used in the neutralisation reactions is …………
Answer:
Normality

Question 21.
The concentration term is used in the calculation of vapour pressure of solution is …………..
Answer:
Mole fraction

Question 22.
The term used to express the active ingredients present in therapeutics is ………
Answer:
Percentage units

Question 23.
When maximum amount of solute is dissolved in a solvent at a given temperature, the solution is called ………..
Answer:
Saturated solution

Question 24.
The solvent in which sodium chloride readily dissolves is …………
Answer:
Water

Question 25.
………… is used by deep-sea divers.
Answer:
Helium, nitrogen and oxygen

Question 26.
The mathematical expression of Raoult’s law is ………..
Answer:
P_A = P_A⁰ . X_A

Question 27.
……….. is an ideal solution?
Answer:
Chloro benzene & bromo benzene

Question 28.
………….. is important in some vital biological systems.
Answer:
osmotic pressure

Question 29.
………. is not a colligative property.
Answer:
vapour pressure

Question 30.
According to van’t Hoff equation. the value of osmotic pressure t is equal to …………
Answer:
π = CRT

Question 31.
The osmotic pressure of the blood cells is approximately equal to at 37°C.
Answer:
7 atm.

Question 32.
Which one of the following is applied in water purification?
Answer:
reverse osmosis

Question 33.
In commercial reverse osmosis process, the semi permeable membrane used is ………..
Answer:
cellulose acetate

Question 34.
The degree of dissociation α is equal to ……….
Answer:
\(\frac { i – 1 }{ n – 1 }\)

Question 35.
The degree of association a is equal to ……….
Answer:
\(\frac { (i – 1)n }{ n – 1 }\)

Question 36.
The estimated vantt Hoff factor for acetic acid solution in benzene is ………..
Answer:
0.5

Question 37.
The estimated van’t Hoff factor for sodium chloride in water is ………..
Answer:
2

Question 38.
Number of moles of the solute dissolved per dm3 of solution is ……….
Answer:
molarity

Question 39.
Molarity of pure water is ………….
Answer:
55.55
Solution:

Question 40.
18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to ………..
Answer:
0.1
Solution:
\(\frac { P° – P }{ P° }\) = x₂
x₂ = No. of moles of glucose
\(\frac { 18 }{ 180 }\) = 0.1
\(\frac { P° – P }{ P° }\) = 0.1

Question 41.
When NaCl is dissolved in water, boiling point ………..
Answer:
increases

Question 42.
Use of glycol as antifreezer in automobile is an important application of …………….
Answer:
Colligative property

Question 43.
Ethylene glycol is mixed with water and used as antifreezer in radiators because …………..
Answer:
it lowers the freezing point of water

Question 44.
Colligative properties of a solution depend on ………… present in it.
Answer:
Number of solute particles

Question 45.
Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ………….
Answer:
low atmospheric pressure

IV. Choose the odd one out.

Question 1.
(a) Air
(b) Camphor in nitrogen gas
(c) Humid oxygen
(d) Salt water
Answer:
(d) Salt water.
a, b and e are gaseous solution whereas d is a liquid solution.

Question 2.
(a) CO₂ dissolve in water
(b) Salt water
(c) Solution of H₂ in palladium
(d) Ethanol dissolved in water
Answer:
(c) Solution of H₂ in palladium
a, b and d are liquid solutions whereas c is a solid solution.

Question 3.
(a) Amalgam of potassium
(b) Camphor in nitrogen gas
(c) Solution of H₂ in palladium
(d) Gold alloy
Answer:
(b) Camphor in nitrogen gas
a, b and d arc solid solutions whereas b is gaseous solution.

Question 4.
(a) Vapour pressure
(b) Lowering ofvapour pressure
(c) Osmotic pressure
(d) Elevation of boiling point
Answer:
(a) Vapour pressure
b, e and dare colligative properties whereas a is a physical property.

Question 5.
(a) Benzene and tolucne
(b) Chlorobenzene and Bromobenzene
(c) Benzene and acetone
(d) n – hexane and n – heptane
Answer:
(a) Benzene and acetone
a, b and dare ideal solutions whereas c is a non-ideal solution.

Question 6.
(a) Ethyl alcohol and cyclohexane
(b) Ethyl bromide and ethyl iodide
(c) Acetone and ethyl alcohol
(d) Benzene and acetone
Answer:
(a) Ethyl bromide and ethyl iodide
a, e and dare non-ideal solutions whereas b is an ideal solution.

V. Choose the correct pair.

Question 1.
(a) Humid oxygen – Liquid solution
(b) Gold alloy – Solid solution
(c) Salt water – Gaseous solution
(d) Solution of H₂ in palladium – Gaseous solution
Answer:
(b) Gold alloy – Solid solution

Question 2.
(a) Air – Gaseous solution
(b) Amalgam of potassium – Liquid solution
(c) Salt water – Solid solution
(d) Carbonated water – Solid solution
Answer:
(a) Air – Gaseous solution

Question 3.
(a) Benzene and toluene – Non-ideal solution
(b) Benzcnc and acetone – Non-ideal solution
(c) Chlorobenzene and bromo henzene – Non-ideal solution
(d) Carbon tetrachloride and Chloroform – ideal solution
Answer:
(b) Benzene and acetone – Non-ideal solution

Question 4.
(a) Benzene and toluene – Ideal solution
(b) n-hexane and n-heptane – Non-ideal solution
(c) Ethyl iodide and ethyl bromide – Non-ideal solution
(d) Chiorobenzene and bromo benzene – Non-ideal solution
Answer:
(a) Benzene and toluene – Ideal solution

VI. Choose the incorrect pair.
Question 1.

Answer:

Question 2.
(a) Benzene and acetone – Ideal solution
(b) Ethyl alcohol and cyclohexane – Non-ideal solution
(C) n-hexane and n-heptanc – Ideal solution
(d) Chioro benzene – Ideal solution
Answer:
(a) Benzene and acetone – Ideal solution

VII Assertion & Reason.

Question 1.
Assertion (A) : When NaCI is added to water, a depression in freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes the depression in freezing poi nl.
(a) Assertion and Reason are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Assertion and Reason are correct and R is the correct explanation of A.

Question 2.
Assertion (A): Ammonia reacts with water does not obey Henry’s law.
Reason (R): The gases reacting with the solvent does not obey Henry’s law.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct hut (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A): Acetic acid solution deviates from Raoult’s law.
Reason (R): Association of solute molecules exists as a dimer by forming intermolecular. hydrogen bonds and hence deviates from Raoult’s law.
(a) Both (A) and (R) arc wrong.
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

VIII. Choose the correct statement.

Question 1.
(a) Raoult’s law is applicable to volatile solid solute in liquid solvent
(b) Henry’s law is applicable to solution containing solid solute in liquid solvent
(c) For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law.
(d) For saturated solution containing volatile solid solute in liquid solvent both laws are obeyed.
Answer:
(c) For very dilute solutions. the solvent obeys Raoult’s law and the solute obeys LIenrys law.

Samacheer Kalvi 11th Chemistry Solutions 2 Marks Questions and Answers

I. Write brief answer to the following questions:

Question 1.
What is the common property observed in naturally existing solution? Explain it.
Answer:

1. Sea water, air are the naturally existing homogeneous mixture. The common property observed in these is homogeneity.
2. The homogeneity implies uniform distribution of their constituents or components through out the mixture.

Question 2.
Define solution with an example.
Answer:
1. A solution is a homogeneous mixture of two or more substances consisting of atoms. ions or molecules.

2. For example, when a small amount of NaCl is dissolved in water, a homogeneous solution is obtained. In this solution, Na⁺ and C^– ions are uniformly distributed in water. Here NaCI is the solute and water is the solvent.

Question 3.
What are aqueous and non aqueous solution? Give example.
Answer:

1. If the solute is dissolved in the solvent water, the resultant solution is called as an aqueous solution. e.g., salt in water.
2. If the solute is dissolved in the solvent other than water such as benzene, ether, CCl₄ etc, the resultant solution is called a non aqueous solution. e.g., Br, in CCI₄.

Question 4.
Define molality.
Answer:
Molality is defined as the number of moles of solute present in 1 kg of the solvent.

Question 5.
Define molaritv.
Answer:
Molarity is defined as the number of moles of solute present in 1 litre of the solution.

Question 6.
Define normality.
Answer:
Normality is deflncd as the number of gram equivalents of solute present in 1 litre of the solution.

Question 7.
Define forniality.
Answer:
Formality (F) is defined as the number of formula weight of solute present in 1 litre of the solution.

Question 8.
Define mole fraction.
Answer:
Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all components present in the solution.

Question 9.
Show that the sum of mole fraction of a solution is equal to one.
Answer:
Consider a solution containing two components A and 13 whose mole fractions are x_A and x_B respectively. Let the number of moles of two components A and B are n_A and n_B respectively.

Question 10.
Define mass percentage.
Answer:
Mass percentage is defined as the ratio of the mass of the solute in g to the mass of solution in g multiplied by 100.

Question 11.
Define volume percentage.
Answer:
Volume percentage is defined as the ratio of volume of solute in mL to the volume of solution in ml multiplied by 100.

Question 12.
Define mass by volume percentage.
Answer:
It is defined as the ratio of the mass of the solute in g to the volume of the solution in ml multiplied by 100.

Question 13.
What is meant by ppm? Where is it used?
Answer:
1. part per million =

2. ppm is used to express the quantity of solutes present in small amounts in solutions.

Question 14.
What is meant by stock solution (or) standard solution? What is meant by working standard?
Answer:
1. A standard solution or a stock solution is a solution whose concentration is accurately known.

2. At the time of experiment, the solution with required concentration is prepared by diluting the stock solution. This diluted solution is called working standard.

Question 15.
Define solubilitv.
Answer:
The solubility of a substance is defined as the amount of the solute that can be dissolved in loo g of the solvent at a given temperature to form a saturated solution.

Question 16.
Ammonia is more soluble than oxygen in water. Why?
Answer:
Ammonia forms hydrogen bonding with water molecules, this intermolecular bonds arc very strong and thus the ammonia is more soluble in water. Ammonia is strongly interact with water to form ammonium hydroxide. But oxygen is more electronegative it is not able to interact with water more. So NH₃ is more soluble than O₂ in water.

Question 17.
Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement.
Answer:
When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.

Question 18.
Dissolution of ammonium nitrate increases with increase in temperature. Why?
Answer:
The dissolution process of ammonium nitrate is endothermic. So the solubility increases with increase in temperature.

Question 19.
What is the relationship between the solubility of eerie sulphate with temperature?
Answer:
The dissolution of eerie sulphate is exothermic and the solubility decreases with the increase in temperature.

Question 20.
Why in the dissolution of CaCl₂, the solubilit increases moderately with high temperature?
Answer:
Even though the dissolution of CaCI₂, is cxothcrmic, the soluhility increases moderately with increase in temperature. Here the entropy factor plays a significant role in deciding the position of equilibrium.

Question 21.
Why the carbonated drinks are stored in pressurized container?
Answer:
1. The carbonated beverages contain CO₂ dissolved in them. To dissolve the CO₂ in these drinks, CO₂ gas is bubbled through them under high pressure.

2. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO₂ drops to the atmospheric pressure level and hence bubbles of CO₂ rapidly escape from the solution and show effervescence.

Question 22.
Define

1. Evaporation
2. Condensation.

Answer:
1. Evaporation:
If the kinetic energy of molecules in the liquid state overcomes the intermolecular force of attraction between them, then the molecules will escape from the liquid state. This process in called evaporation.

2. Condensation:
The vapour molecules are in random motion during which they collide with each other and also with the walls of the container. As the collision is inelastic, they lose their energy and as a result the vapour returns back to liquid state. This process is called as condensation.

Question 23.
State Dalton’s law of partial pressure.
Answer:
According to Dalton’s law of partial pressure, the total pressure in a closed vessel will be equal to the sum of the partial pressure of the individual components.
P_total = P_A + P_B

Question 24.
Give the reason behind the lowering of vapour pressure in the dissolution of NaCl in water?
Answer:
NaCI is a non volatile solute. When a non volatile solute is dissolved in pure solvent, the vapour pressure of pure solvent will decrease. In such solution, vapour pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.

Question 25.
What are ideal solution? Give example.
Answer:
An ideal solution is a solution in which each component i.e., the solute as well as the solvent obeys the Raoult’s law over the entire range of concentration.

Question 26.
What are non-ideal solution? Give example.
Answer:

1. The solutions which do not obey Raoult’s law over the entire range of concentration are called non-ideal solutions.
2. The deviation of the non-ideal solution from the Raoult’s law may be positive (or) negative.
3. Example, Ethyl alcohol and cyclohexane.

Question 27.
What are colligative properties? Give example.
Answer:
The properties which do not depend on the chemical nature of the solute but depends only on the number of solute particles present in the solution are called colligative properties. e.g.,

1. Relative lowering of vapour pressure – \(\frac { P° – P}{ P° }\)
2. Osmotic pressure – π
3. Elevation of boiling point – ∆T_b
4. Depression in freezing point – ∆T_f

Question 28.
What is meant by elevation of boiling point?
Answer:
1. The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure.

2. When a non-volatile solute is added to pure solvent at its boiling point, the vapour pressure of the solution is lowered below 1 atm. To bring the vapour pressure again to 1 atm, the temperature of the solution has to be increased.

3. As a result, the solution boils at a higher temperature (T_b) then the boiling point of pure solvent (T_b°). This increase in the boiling point is known as elevation of boiling point.

Question 29.
Define ebullioscopic constant.
Answer:
Ebullioscopic constant k_b, is equal to the elevation in boiling point for 1 molal solution.

Question 30.
Define osmotic pressure.
Answer:
Osmotic pressure can be defined as the pressure that must be applied to the solution to stop the influx of the solvent (to stop osmosis) through the semipermeable membrane.

Question 31.
Write the Van’t Hoff equation of osmotic pressure.
Answer:
Van’t Hoff equation states that for dilute solutions, the osmotic pressure is directly proportional to the molar concentration of the solute and the temperature of the solution.
π = CRT
where
π = Osmotic pressure
C = concentration
T = Temperature
R = gas constant

Question 32.
Define Van’t Hoff factor.
Answer:
van’t Hoff factor (I) is defined as the ratin of the actual molar mass to the abnormal molar mass of the solute.

Question 33.
How is degree of dissociation and degree of association are related with van’t Hoff factor?
Answer:
The degree of dissociation or association can be related to van’t Hoff factor
1. using the following relationship

• α_dissociation = \(\frac { i – 1 }{ n – 1 }\)
• α_association = \(\frac { (1 – i)n }{ n – 1 }\)

where n = number of solute particles

Question 34.
Give an example of a solid solution ¡n which the solute is a gas.
Answer:
Solution of hydrogen in palladium.

Question 35.
What role does the molecular interaction play in solution of alcohol and water?
Answer:
There is strong hydrogen bonding in alcohol molecules as well as water molecules. The intermolecular forces both in alcohol and water are H-bonds. When alcohol and water are mixed,

they form solution because of formation of H-bonds between alcohol and H₂O molecules hut these interactions are weaker and less extensive than those in pure water. Hence, they show positive deviation from ideal behaviour.

Question 36.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. it is because the fact that this process involves decrease of entropy. Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.
(Gas + Solvent \(\rightleftharpoons\) Solution + Heat)

Question 37.
Why is the freezing point depression of 0.1 M NaCl solution nearly twice that of 0.1M glucose solution?
Answer:
NaCl is an electrolyte and it dissociates completely whereas glucose being a non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double for NaCI solution than that for glucose solution of same molarity.

Therefore depression in freezing point being a colligative property ¡s nearly twice for NaCl solution than that for glucose solution of same molarity.

Question 38.
Why a person suffering from high blood pressure is advised to take minimum quantity of common salt?
Answer:
Osmotic pressure is directly proportional to the concentration of solutes. Our body fluid contains a number of solutes. On taking large amount of salt, ions entering into the body fluid thereby raises the concentration of solutes. As a result, osmotic pressure increases which may rupture the blood cells.

Samacheer Kalvi 11th Chemistry Solutions 3 Marks Questions and Answers

Question 1.
What are gaseous solution ? Give its various types with example
Answer:

Question 2.
What are liquid solutions ? Explain with example
Answer:

Question 3.
What are solid solution? Give example.
Answer:

Question 4.
How will you prepare a standard solution?
Answer:

1. A standard solution or a stock solution is a solution whose concentration is accurately known.
2. A standard solution of required concentration can be prepared by dissolving a required amount of a solute in a suitable amount of solvent.
3. It is done by transforming a known amount of solute to a standard flask of definite volume. A small amount of water is added lo the flask and shaken well to dissolve the salt.
4. Then water is added to the flask to bring the solution level lo the mark indicated at the top end of the flask.
5. The flask is stoppered and shaken well to make concentration uniform.

Question 5.
What are the advantages of standard solution.
Answer:
1. The error due to weighing the solute can be minimised by using concentrated stock solution that requires large quantities of solute.

2. We can prepare working standards of different concentrations by diluting the stock solution which is more efficient since consistency is maintained.

3. Some of the concentrated solutions are more stable and are less likely to support microbial growth than working standards used in the experiments.

Question 6.
Explain the solubilities of ammonium nitrate, calcium chloride, ceric sulphate and sodium chloride in water at different temperature with a graph.
Answer:
1. The solubility of sodium chloride does not vary appreciably as the maximum solubility is achieved at normal temperature. In fact, there is only 10% increase in solubility between 0°C to 100°C.

2. The dissolution process of ammonium nitrate is endothermic, the solubility increases with

3. In the case of eerie sulphate. the dissolution is exothermic and the solubility decreases with increase in temperature.

4. Even though the dissolution of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here the entropy factor also plays a significant role in deciding the position of equilibrium.

Question 7.
Explain the effect of temperature gaseous solute ¡n liquid solvent.
Answer:
1. In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature.

2. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak inter molecular forces when the temperature increases, the average. kinetic energy of the molecules present in the solution also increases.

3. The increase in kinetic energy breaks (he weak inter molecular forces between the gaseous solute and liquid solvent with results in the release of the dissolved gas molecules to gaseous state.

4. The dissolution of most of the gases in Liquid solvents is an endothermic process, the increase in temperature decreases the dissolution of gaseous molecules.

Question 8.
Give reason why aquatic species are less sustained in hot water?
Answer:
There will be decrease in solubility of gases in solution with increase in temperature. During summer, in hot water rivers, due to high temperature. the availability of dissolved oxygen decreases. So the aquatic species are less sustained in hot water.

Question 9.
Deep – sea divers use air diluted with helium gas in their tanks. Why? (or) Justify this statement.
Answer:
1. Deep-sea divers carry a compressed air tank for breathing at high pressure under water. This air tank contains nitrogen and oxygen which are not very soluble in blood and other body fluids at normal pressure.

2. As the pressure at the depth is far greater than the surface atmospheric pressure, more nitrogen dissolves in the blood when the diver breathes from tank.

3. When the divers ascends to the surface, the pressure decreases, the dissolved nitrogen comes out of the blood quickly forming bubbles in the blood stream.

These bubbles restrict blood flow, affect the transmission of nerve impulses and can even burst the capillaries or block them. This condition is called “the bends” which are painful and dangerous to life.

4. To avoid such dangerous condition they use air diluted with helium gas (11.7 % helium, 56.2% nitrogen and 32.1% oxygen) of lower solubility of helium in the blood than nitrogen.

Question 10.
What are the limitations of Henry’s law?
Answer:

1. Henry’s law is applicable at moderate temperature and pressure only.
2. Only the less soLuble gases obey Henry’s law.
3. The gases reacting with solvent do not obey Henry’s law.
4. The gases obeying Henrys law should not associated or dissociated while dissolving in the solvent.

Question 11.
Explain how benzene in toluene obeys Raoult’s law.

Answer:
The variation of vapour pressure of pure benzene and toluenc with its mole fraction is given in the graph.
1. The vapour pressure of pure toluene and pure benzene are 22.3 and 74.7 mm Hg respectively.

2. The graph shows the partial vapour pressure of pure components increases linearly with the increase of the mole fraction of the respective components. The total pressure at any composition of the solute and solvent is given by the straight line.

3. P_solution = P⁰_toluene + x_benzene (P⁰_benzene – P⁰_toluene)

Question 12.
Derive the relationship between the relative lowering of vapour pressure and mole fraction of the solute.
Answer:
P_solution ∝ x_A by Raoult’s law. where x_A is the mole fraction of the solvent.
P_solution = k . x_A
When
x_A = 1
k = P⁰_solvent
P⁰_solvent = partial pressure of pure solvent
P_solution = P⁰_solvent . x_A

where
x_B = mole fraction of the solute
x_A + x_B = 1
x_B = 1 – x_A

Question 13.
How would you compare Raoult’s law and Henry’s law.
Answer:
1. According to Raou It’s law, for a solution containing a non volatile solute.
P_solution = P⁰_solvent . x_solute

2. According to henry’s law, P_solution = K_H . x_solute in solution

3. The difference between the above two laws is the proportionality constant P° (Raoult’s law) and K_H (Heniys law).

4. henry’s law is applicable to solution containing gaseous solute in liquid solvent, while Raoult’s law is applicable to non volatile solid solute in the liquid solvent.

5. If the solute is non volatile then the Henry’s law constant will become equal to the vapour pressure of pure solvent Po. thus Raoult’s law becomes a special case of Henry’s law.

6. For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law.

Question 14.
What are the necessary conditions for an ideal solution? Give two example. For an ideal solution
1. There is no change in volume on mixing two components (solute and solvent)
∆V_mixing = O

2. There is no exchange of heat when the solute is dissolved in solvent (∆H_mixing = 0)

3. Escaping tendency of the solute and the solvent present in it should be same as in pure liquids.

4. Examples – For ideal solution: Benzene and toluene, n-Hexane and n-Heptane, ethyl bromide and ethyl iodide, chlorobenzene and bromo benzene.

Question 15.
Explain how non-ideal solutions shows positive deviation from Raoult’s law.
Answer:

1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water.
2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and water-water interaction).
3. This results in the increased evaporation of both componeins from the aqueous solution of ethanol.
4. Consequently, the vapour pressure of the solution is greater than the vapour pressure predicted by Raoult’s law.
5. Here, the mixing OCCSS is endothermic i.e., (∆H_mixing > O) and there will be a slight increase in volume (∆V_mixing > O)

Question 16.
Explain with suitable example about negative deviation from law.
Answer:
1. Let us consider a solution of phenol and aniline. Both phenol and aniline form hydrogen bonding interactions amongst themselves.

2. When mixed with aniline, the phenol molecule forms hydrogen bonding interactions with aniline, which are Stronger than the hydrogen bonds formed amongst themselves.

3. Formation of new hydrogen bonds considerably reduce the escaping tendency of phenol and aniline from the solution.

4. Asa result, the vapour pressure of the solution is less and there is a slight decrease in volume (∆V_mixing < 0) on mixing.

5. During this process evolution of heat takes place i.e., ∆V_mixing < 0 (exothermic).

6. Examples – Acetone + Chloroform, Chloroform + Diethyl ether

Question 17.
The vapour pressure of a solution containing a non volatile, non-electrolyte solute is always lower than that of pure solvent. Give reason.
Answer:
1. The vapour pressure of a solution (P) containing flOfl volatile solute is lower than that of pure solvent (P°).

2. Consider a closed system is which a pure solvent is in equilibrium with its vapour. At equilibrium the molar Gibbs free energies of solvent in a liquid and gaseous phase are equal (∆G = O).

3. When a solute is added to this solvent the dissolution takes place and its free energy (G) decreases due to increase in entropy.

4. In order to maintain the equilibrium, the free energy of the vapour phase must also decrease.

5. At a given temperature, the only way to lower the free energy of the vapour is to reduce its pressure.

6. Thus the vapour pressure of the solution must decrease to maintain the equilibrium.

Question 18.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
According to Raoult’s law,
P_solution x_A, where x_A = mole fraction of solvent.
P_solution = k . x_A, where k = proportionality constant
For a pure solvent,
Vapour pressure = P°, x_A = 1
P°_solution = k x 1 = k
Substituting P°_solvent in Raoult’s law
P_solution = P°_solvent . x_A
Relative lowenng of vapour pressure

substituting P_solution as P°x_B in the above eaquation
Relative lowering of vapour pressure

x_A + x_B = I
where x_B = mole fraction of solute. It is clear that the relative lowering of vapour pressure depends only on the mole fraction ofthe solute (x_B) and is independent of its nature. Therefore relative lowering of vapour pressure is a colligative property.

Question 19.
Explain why boiling point of solution is greater than that of pure solvent?
Answer:
When a non volatile solute is added to a pure solvent at its boiling point, the vapour pressure of the solution is lowered below 1 atm. To bring the vapour pressure again to I atm the temperature of the solution has to be increased.

As a result, the solution boils at a higher temperature (T_b) than the boiling point of the pure solvent (T°_b). This increase in the boiling point is known as elevation of boiling point ∆T_b = T_b – T°_b.

Question 20.
Graphically prove that Tb ¡s greater than T°_b.
Answer:

1. The vapour pressure of the solution increases with increase in temperature. The variation of vapour pressure with respect to temperature of pure water is given by the curve – A.

2. At 100°C, the vapour pressure of water is equal to I atm. Hence, the boiling point of water is 100°C (T°_b).

3. When a solute is added to water, the vapour pressure of the resultant solution is lowered. The variation of vapour pressure with respect to temperature for the solution is given by curve-B.

4. From the graph, it is evident that the vapour pressure of the solution is equal to 1 atm. pressure at the temperature T_b which is greater than T°_b. The difference between these two temperatures (T_b – T°_b) gives the elevation of boiling point.
∆T_b = T_b T°_b.

Question 21.
Derive the relationship between the elevation of boiling point and molar mass of non volatile solute.
Answer:
The elevation of boiling point ∆T_b = T_b T°_b.
∆T_b is directly proportional to the concentration of the solute particles.
∆T_b ∝ m, (m = molaLiiy)
∆T_b = k_b. m, where k_b = ebullioscopic constant

Question 22.
Define

1. freezing point
2. Depression in freezing point.

Explain with graph.
Answer:

1. Freezing point is defined as the temperature at which the solid and the liquid states of the substances have the same vapour pressure.

2. When a non volatile solute is added to water at its freezing point, the freezing point of water is lowered from 0°C. The lowering of freezing point of the solvent when a solute is added is called depression in freezing point AT1..

3. ∆T_f = T⁰_f – T_f

Question 23.
Define

1. cryoscopic constant
2. ebullioscopic constant

Answer:

1. ∆T_f = k_f. m, where k_f = cryoscopic constant. If m = 1. then AT_f = k_f
k_f is defined as depression in freezing point for 1 molal solution.

2. ∆T_f = k_f. m where k_f ebullioscopic constant. If m = 1 then AT_f = k_f
k_b is defined as elevation in boiling point for 1 molal solution.

Question 24.
What are the significances of osmotic pressure over other colligative properties ?
Answer:
1. Unlike elevation of boiling point and the depression in freezing point, the magnitude of osmotic pressure is large.

2. The osmotic pressure can be measured at room temperature enables to determine the molecular mass of biomolecules which are unstable at higher temperature.

3. Even for a very dilute solution, the osmotic pressure is large.

Question 25.
What is haemolysis ? intravenous fluid are isotonic to blood?
Answer:
1. The osmotic pressure of the blood cells is approximately equal to 7 atm at 37°C.

2. The intravenous injections should have saine osmotic pressure as that of the blood (isotonic vith blood).

3. If the intravenous solutions are too dilute that is hypotonie, the solvent from outside of the cells flow into the cell to normalise the osmotic pressure and this process is called haernolysis causes the cells to burst.

4. On the other hand, if the solution is too concentrated, that is hypertonic. the solvent molecules will flow out of the cells,which causes the cells to shrink and die.

5. For this reason, the intravenous fluids are prepared such that they are isotonic to blood (0.9% mass/volume sodium chloride solution).

Question 26.
Explain reverse osmosis.
Answer:
1. The pure water moves through the semipermeable membrane to the NaCl solution due to osmosis.

2. This process can be reversed by applying pressure greater than the osmotic pressure to the solution side. Now the pure water moves from the solution side to the solvent side and this process is called reverse osmosis.

3. Reverse osmosis can be defined as a process in which a solvent passes through a semipermeable membrane in the opposite direction of osmosis, when subjected to a hydrostatic pressure greater than the osmotic pressure.

Question 27.
Explain about the application of reverse osmosis in water purification.
Answer:
1. Reverse osmosis is used in the desalination of sea water and also in the purification of drinking water.

2. When a pressure higher than the osmotic pressure is applied on the solution side (sea water) the water molecules moves from solution side to the solvent side through semi permeable membrane (opposite to osmotic flow). The pure water can be collected.

3. Cellulose acetate (or) polyamide membranes are commonly used in commercial system.

Question 28.
Acetic acid is found to have molar mass as 120 g mol^-1. Prove it.
Answer:
1. In certain solvent, solute molecules associate to form a dimer. This reduces the total number of molecules formed in solution and as a result the calculated molar mass will be higher than the actual molar mass.

2. Acetic acid in benzene exist as a dimer

3. The molar mass of acetic acid calculate using colligative properties is found to be around 120 g mol^-1 is two times of the actual molar mass 60 g mol^-1.

Question 29.
Depression in freezing point of NaCI is twice that of in urea. Why?
Answer:
1. The electrolyte NaCI dissociates completely into its constituent ions in their aqueous solution. This causes an increase in the total number olparticles present in the solution.

2. When we dissolve 1 mole of NaCI in water it. dissociates and gives 1 mole of Na⁺ and 1 mole of Cl^–. Hence the solution will have 2 moles of particles.

But when we dissolve 1 mole of urea (non electrolyte) in water it appears as 1 mole only. So the colligative property value would be double in NaCl than in urea.

Question 30.
What is van’t Hoff factor? Calculate the van’t Hoff factor value for

1. acetic acid
2. NaCl

Answer:
1. van’t Hoff factor is defined as the ratio ofthe actual molar mass to the abnormal (calculated) molar mass of the solute.

2.

van’t Hoff factor (1) for acetic acid = \(\frac { 60 }{ 120 }\) = 0.5

3. van’t Hoff factor (2) for NaCl = \(\frac { 117 }{ 58.5 }\) = 2

Question 31.
Differentiate between ideal solution and non-ideal solution.
Answer:
Ideal solution
An ideal solution is a solution in which each component obeys the Raoult’s law over the entire range of concentration.
For an ideal solution,

• ∆H_mixing = 0
• ∆V_mixing = 0

Example: Benzenc and toluene n – Hexane and n – Heptane

Non-ideal solution
The solutions which do not obey Raoult’slaw over the entire range of concentrationsare called non-ideal solution.
For a non-ideal solution.

• ∆H_mixing \(\quad \neq\) 0
• ∆V_mixing \(\quad \neq\) 0

Example: Ethyl alcohol and Cyclo hexane, Benzene and acetone .

Question 32.
Explain the factors when i = 1, i < 1 and i >1 ?
Answer:
1. For a solute that does not dissociate or associate the vant’s hoff factor is equal to 1 (i = 1) and the molar mass will be close to the actual molar mass.

2. For that solute that associate to form higher oligomers in solution, the van’t Hoff factor will be less than 1 (i < 1) and the observed molar mass will be greater than the actual molar mass.

3. For solutes that dissociates into their constituent ions the van’t Hoff factor will be more than one (i > 1) and the observed molar mass will be less than the normal molar mass.

Question 33.
State Henry’s law and mention some of its important applications.
Answer:
Henry’s law: The solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Application of Henry’s law:

1. In the production of carbonated beverages
   (as solubility of CO₂ increase at high pressure).
2. In the deep sea diving.
3. In the function of lungs.
4. For climbers or people living at high altitudes,

Question 34.
What type of non – idealities are exhibited by cyclohexane – ethanol and acetone – chloroform mixture? Give reason for your answer.
Answer:
Ideal solutions are those which obey Raoult’s law over extreme range of concentration. Ideal solutions have another important properties:

• ∆H_mix = 0
• ∆V_mix = 0

Here-forces of attraction between A – A. B – B and A – B are of the same order. Non ideal solutions do not obey Raoult’s law over the entire range of concentration.
∆H_mixing\(\quad \neq\) 0 and ∆V_mixing\(\quad \neq\) 0

Cyclohexane – ethanol mixture shows positive deviation from Raoult’s law because forces of attraction between cyclohexane and ethanol are less than in between pure cyclohexane as well as pure ethanol.

Acetone-Chloroform mixture shows negative deviation from Raoults law because forces of attraction between acetone and chloroform are higher than that in between pure acetone and pure chloroform molecules.

Question 35.
Given below is the sketch of a plant for carrying out a process.

1. Name the process occurring in the above plant.
2. To which container does the net flow of solvent take place?
3. Name one SPM which can he used in this plant.
4. Give one practical use of the plant.

Answer:

1. Reverse osmosis
2. In fresh water container from salt water container.
3. Cellulose acetate is semipermeable membrane (SPM)
4. Purification of water

Question 36.
Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure?
Answer:
π = CRT
π = \(\frac { n }{ V }\)RT
πV = nRT

Osmotic pressure is inversely proportional to the molecular mass of the soLute.

Question 37.
1. Menthol is a crystalline substance with peppermint taste. A 6.2% solution of menthol in cyclohexane freezes at – 1.95°C.

Determine the formula mass of menthol. The freezing point and molal depression constant of cyclohexane are 6.5°C and 20.2 K m^{-1, respectively.}

2. State Henry’s Law and mention its two important applications.

3. Which of the following has higher boiling point and why? 0.1 M NaCl or 0.1 M Glucose
Answer:
1.

M_B = 158 g mol^-1

2. Henry’s Law:
The solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Applications:

• Solubility of CO₂ is increased at high pressure.
• Mixture of He and O₂ are used by deep sea divers because he is less soluble than nitrogen.

3. 0.1 M NaCI, because it dissociates in solution and furnishes greater number of particles per unit volume while glucose being a non-electrolyte does not dissociate.

Question 38.
Water is a universal solvent. But alcohol also dissolves most of the substances soluble in water and also many more. Boiling point of water is 100°C and that of alcohol is 80°C. The specific heat of water is much higher than the specific heat of alcohol.

1. List out three possible differences if instead of water as the liquid ¡n our body we had alcohol.
2. What value can you derive from this special property of water and its innumerable uses in sustaining life on earth?

Answer:
1.
(i) Even a small rise in temperature in the surroundings will raise the temperature of’ the body because the specific heat of alcoholis much less than the specific heat of water. So, in order to cool the body, more sweating will take place.

(ii) As there is less H bonding in alcohol, it will gel evaporated faster. The alcohol will be evaporated at such a faster rate that the liquid has to be ingested all the time.

(iii) Ice which floats on water helps aquatic life to exist even in winter as water insulates the heat from liquid below it to go back to the surroundings. Solid alcohol does not have such special properties.

2. Praise is to the almighty that has so thoughtfully given such special properties to water and made it a liquid that could sustain life on earth.

Question 39.
State Henry’s law for solubility of a gas in a liquid. Explain the significance of Henry’s law constant (K_H). At the same temperature, hydrogen is more soluble in water than helium. Which of theni will have a higher value of K_H and Why?
Answer:
Henry’s law states that the solubility of a gas in liquid at a given temperature is directly proportional to the partial pressure of the gas.
P = K_H x
where P is the pressure of the gas, x is the mole fraction of the gas in the solution and K_H is the Henry’s law constant. KH is a function of the nature oIgas.

Higher the value of K_H at a given temperature. lower is the solubility of the gas in the liquid. As helium is less soluble in water, so it has a higher value of K_H than hydrogen.

Henry’s Law:
As dissolution of agar in liquid is an exothermic process, therefore, the solubility should decrease with in increase in temperature.

Question 40.
What is meant by positive and negative deviations from Raoult’s law and how is the sign ∆H_mix of related to positive and negative deviations from Raoult’s law?
Answer:
Negative deviations:
In these type of deviations, the partial vapour pressure of each component A and B of solution is higher than the vapour pressure calculated from Raoult’s law. For example -Water and ethanol, chloroform and water.

Positive deviations:
In case of positive deviation A – B interactions are weaker than those between A – A or B – B. This means that in such solutions molecules or A (or B) will find it easier to positive deviation from Raoult’s law.

Samacheer Kalvi 11th Chemistry Solutions 5 Marks Questions and Answers

II. Answer the following questions in detail:

Question 1.

1. Define solution.
2. Explain the types of solutions with suitable example.

Answer:
1. A solution is a homogeneous mixture of two or more substances, consisting of atoms. ions or molecules.

2. Types and examples of solution.

Question 2.

1. Define Solubility
2. Explain about the factors that influences the solubility

Answer:

1. The solubility of a substance at a given temperature is defined as the amount of the solute that can be dissolved in 100 g of the solvent at a given temperature to form a saturated solution.

2. Factors influencing solubility
(a) Nature of solute and solvent: Sodium chloride, an ionic compound readily dissolves in polar solvent such as water but it does not dissolve in non polar solvent such as benzene. Most of the organic compounds dissolve in organic solvent and do not dissolve in water.

(b) Effect of temperature: Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. The dissolution of NaCl does not vary as the maximum solubility is achieved at normal temperature.

The dissolution of ammonium nitrate is endothermic, the solubility increases with increase in temperature. The dissolution of eerie sulphate is exothermic and the solubility decreases with increase of temperature. In the case of gaseous solute in liquid solvent the soluhility decreases with increase in temperature.

Effect of pressure:
Generally the change in pressure does not have any significant effect in the solubility of solids and l?quids as they are not compressible. However, the solubility of gases generally increases with increase of pressure.

Question 3.
Explain about the factors that are responsible for deviation from Raoult’s law.
Answer:
1. Solute-solvent interactions:
For an ideal solution, the interaction between the solvent molecules (A – A), the solute molecules (B – B) and between the solvent and solute molecules (A – B) are expected to be similar. if these interactions are dissimilar, there will be a deviation from ideal behaviour.

2. Dissolution of solute:
When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the solvent and causes deviation from Raoult’s law. e.g., KCI in water deviates from ideal behaviour due to dissociation as K⁺ and Cl^– ion which form strong ion-dipole interaction with water molecules.

3. Association of solute:
Association of solute molecules can also cause deviation from ideal behaviour. For example in solution acetic acid exists as a dimer by forming intermolecular hydrogen bonds and hence deviates from Raoult’s law.

4. Temperature:
An increase in temperature of the solution increases the average kinetic energy of the molecules present in the solution which cause decrease in the attractive force between them. As result, the solution deviates from Raoult’s law.

5. Pressure:
At high pressure, the molecules tends to stay close to each other and therefore there will be an increase in their intermolecular attraction. Thus a solution deviates from Raoult’s law at high pressure.

6. Concentration:
When the concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from Raoult’s law.

Question 4.
How would you determine molar mass from relative lowering of vapour pressure.
Answer:
1. The measurement of relative lowering of vapour pressure can be used to determine the molar mass of a non-volatile solute.

2. A known mass of the solute is dissolved in a known quantity of solvent. The relative lowering of vapour pressure is measured experimentally.

3. According to Raoult’s law, the relative lowering of vapour pressure is

W_A = weight of solvent
W_B = weight of solute
M_A = Molar mass of solvent
M_B = molar mass of solute

where n_A = number of moles of solvent
n_B = number of moles of solute.
For dilute solution, n_A > > n_B, n_A + n_B = n_A
Then,
Number of moles of solvent and solute arc


From the above equation, molar mass of the solute M_B can be calculated using the known values of W_A, W_B, M_A and the measured relative lowering of vapour pressure.

Question 5.
How would you determine the molar mass from osmotic pressure.
Answer:
According to van’t Hoff equation
π = CRT
C = \(\frac { n }{ V }\)
Here n = number of moles of solute dissolved in ‘V’ litre of the solution.
π = \(\frac { n }{ V }\) . RT = πV = nRT
If the solution is prepared by dissolving W_B of the non-volatile solute in W_A g of solvent, then the number of moles of ‘n’ is

where
M_B = molar mass of the solute
Substituting n value, we get

From the above equation, we can calculate the molar mass of the solute.

Question 6.
What are ideal and non-ideal solutions? Explain with suitable diagram the behaviour of ideal solutions.
Answer:
Ideal solutions:
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. Ideal solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecular forces.

Examples:

1. Benzene and toluene
2. n – Hexane and n-Heptane
3. Chiorobenzene and bromobenzene.

Characteristics:

1. They must obey Raoult’s law.
2. ∆H mixing should be zero.
3. ∆W mixing should be zero, i.e. volume change on mixing is zero.

Non – ideal solutions:
The solutions which do not obey Raoult’s law are called non-ideal solutions. In case of non – ideal solutions there is a change in volume and heat energy when the two components are mixed.

Characteristics:

1. They does not obey Raoult’s law.
2. ∆Vmix \(\quad \neq\) 0
3. ∆Hmix \(\quad \neq\) 0

Behaviour of Ideal Solutions:
A plot of P₁ or P₂ versus the mole fraction x₁ and x₂ for an ideal solution gives a linear plot. These Lines (I and II)pass through the points and respectively whenx1 and x₂ is equal to unity.

Similarly the plot (Line III) of P_total versus x₂ is also linear. The minimum value of is P₁° and the maximum value is P₂°, assuming that component 1 is less volatile than component 2, i.e. P₁° < P₂₀.

Question 7.
Explain with a suitable diagram and appropriate example, why some non-ideal solution shows positive deviation from Raou It’s law.
Answer:
Some non-ideal solutions show positive deviation from Raoult’s law. Consider a solution of two components A and B.  If A-B interactions in the solution are weaker than the A – A and B – B interactions in the two liquids forming the solution, then the escaping tendency of molecules A and B from the solution become more than in pure liquids.

The total vapour pressure will be greater than the corresponding vapour pressure as expected on the basis of Raoult’s law. This type of behaviour of solution is called positive deviation from Raoult’s law. The boiling point of such solutions are lowered. Mathematically,
P_A < P_A⁰ x_A
P_B < P_B⁰ x_B
The total vapour pressure is less than P_A + P_B
P < P_A + P_B
P < P°_A x x_A + P°_B x_B
Hence
P₁ = P_A
P₂ = P_B
Examples of solutions showing positive deviations

1. Ethyl alcohol and water
2. Benzene and acetone
3. Ethyl alcohol and cyclohcxanc
4. Carbon tetrachloride and chloroform.

Question 8.

1. What is Osmotic pressure and how is it related to the molecular mass of a non volatile substances?
2. What advantage the osmotic pressure method has over the elevation of boiling point method for determining the molecular mass?

Answer:
1. Osmotic pressure:
It is the pressure of the solution column that can prevent the entry of solvent molecules through a semi-permeable membrane, when the solution and the solvent are separated by the same. It is denoted by π. Its unit is mm 11g or atmosphere.
We know that, π = CRT
where π is the osmotic pressure and R is the gas constant
π = \(\frac { { n }_{ 2 } }{ V }\) RT
where V is volume of solution per litre containing n₂ moles of solute.

By the above relation molar mass of solute can be calculated.

2. The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and molarity of the solution is used instead of molality.

The technique of osmotic pressure for determination of molar mass of solutes is particularly useful for biomolecules as they are generally not stable at higher temperature and polymers have poor solubility.

III. Numerical Problems

Question 1.
Calculate the mole fraction of benzene ¡n solution containing 30% by mass in carbon tetrachioride.
Solution:
30% of benzene in carbon tetrachloride by mass means that Mass of benzene in the solution = 30g
Mass of solution = 100g
Mass of carbon tetrachloride = 100g – 30g = 70g
Molar mass of benzene (C₆H₆) = 78 g mol^-1
Molar mass of CCl₄ = 12 + (4 x 35.5) 154g mol^-1

Question 2.
Calculate (he molarity of each of the following solutions:
Solution:

1. 30 g of CO(NO₃)₂. 6H₂O = in 4.3 L of solution
2. 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

1. Molar mass of CO(NO₃)₂. 6H₂O = 58.7 + 2(14 + 48) + (6 x 8)g mol^-1
= 58.7 + 124 + 108g mol^-1 = 290.7 gmol^-1

Volume of solution = 4.3 L

2. 1000 mL of 0.5 M H₂SO₄ Contain H₂SO₄ = 0.5 moles
30 mL of 0.5 M H₂SO₄ contain H₂SO₄ = \(\frac { 0.5 }{ 1000 }\) x 30 mole = 0.015 mole
Volume of solution = 500 mL = 0.500 L

Question 3.
Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution:
0.25 molal aqueous solution means that
Moles of urea = 0.25 mole
Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea = 14 + 2 + 12 + 16 + 14 + 2 = 6Og mol^-1
0.25 mole of urea = 60 x 0.25 mole = 15 g
Total mass of the solution = 1000 + 15 g = 1015 g = 1.015 g
Thus, 1.015 kg of solution contain urea = 15 g
2.5 kg of solution will require urea = \(\frac { 15 }{ 1.015 }\) x 2.5 kg = 37g

Question 4.
H₂S, a toxic gas with rotten egg like smell is used for the qualitative analysis. 1f the solubility of H₂S in water at STP is 0.195 m. Calculate Henrvs law constant.
Solution:
Solubility of H₂S gas = 0.195 m
= 0.195 mole in 1 kg of the Solvent (water)
1 kg of the solvent (water) = 1000 g
Mole fraction of H₂S gas in the solution (x) = \(\frac { 0.195 }{ 0.195 + 55.55 }\) = 0.0035
Pressure at STP = 0.98 7 bar
Applying Henry’s law
P_H2S = K_H x x_H2S

Question 5.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:
Here
P°₁ = 23.8 mm
W₂ = 50g
M₂ (urea) = 60g mol^-1
W₁ = 850g
M₁(Water) = 18g mol^-1
Here we have to calculate P_s
Applying Raoult’s law,

Thus, relative lowering of vapour pressure = 0.017
Substituting P° = 23.8 mm Hg

We get,
23.8 – P_s = 0.017 P_s
P_s = 23.4 mm Hg
Thus, vapour pressure of water in the solution = 23.4 mm Hg

Question 6.
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500g of water such that it boils at 100°C? .
Solution:
Elevation in boiling point required, ∆T_b = 100 – 99.63° = 0.37°
Mass of solvent (water) W₁ = 500g
Mass of solute, C₁₂H₂₂O₁₁ = 342 g mol^-1
Molar mass of solvent M₁ = 18 g mol^-1
Applying the formula,

Question 7.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504g mL^-1?
Solution:
68% nitric acid by mass means that
Mass of nitric acid = 68 g
Mass of solution = 100 g
Molar mass of HNO₃ = 63g mol^-1
68g HNO₃ = mole = 1.079 mole
Density of solution = 1.504 g mL^-1
Volume of solution = \(\frac { 100 }{ 1.504 }\) mL = 66.5 mL = 0.0665 L

Question 8.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate tile mass percentage of the resulting solution.
Solution:
300 g of 25% solution contains solute = 75g
400g of 40% solution contains solute = 160 g
Total mass of solute = 160 + 75 = 235 g
Total mass oÍ’ solution = 300 + 400 = 700 g
% of solute in the final solution = \(\frac { 235 }{ 700 }\) x 1oo = 33.5%
% of water in the finaI solution = 100 – 33.5 = 66.5%

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCI₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass)

1. express this in percentage by mass
2. determine the molality of chloroform in the water sample.

Solution:
1. 15 ppm means 15 parts in million (10₆) parts by mass in the solution
% of mass = \(\frac { 15 }{ { 10 }^{ 6 } }\) x 100 = 1.5 x 10^-4

2. Taking 15g chloroform in 106g of the solution
Mass of the solvent = 10⁶ g
Molar mass of CHCl₃ = 12 + 1 + (3 x 35.5) = 119.5 g mol^-1

Question 10.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at normal boiling point of the solvent. What is the molar mass of the solute?
Solution:
Vapour pressure of pure water at the boiling point
(P°) = 1 atm 1.013 bar
Vapour pressure of solution P_s = 1.004 bar
M₁ = 18 g mol^-1
M₂ = ?
Mass of solute = W₂ = 2g
Mass of solution = 100g
Mass of solvent W₁ = 98 g
Applying Raoult’s law for dilute solution

Question 11.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K.
Solution:
Molar mass of cane sugar
C₁₂H₂₂O₁₁ = 342 g mol^-1
Molality of sugar = \(\frac { 5 x 1000 }{ 342 x 100 }\) = 0.146
∆T₂ for sugar solution = 273.15 – 271 = 2.15°
∆T_f = K_f x m
K_f = \(\frac { 2.15 }{ 0.146 }\)
Molality of glucose solution = \(\frac { 5 }{ 180 }\) x \(\frac { 1000 }{ 100 }\) = 0.278 m
∆T_f (Glucose) = \(\frac { 2.15 }{ 0.146 }\) x 0.278 = 4.09°K
Freezing point of glucose solution = 273.15 – 4.09 = 269.06 K

Question 12.
Calculate the amount of benzoic acid (C₆HCOOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:
0.15 M solution means that 0.15 moles of C₆H₅COOH is present in IL
= 1000 mL of the solution
Molar mass of C₆H₅COOH = 72 + 5 + 12 + 32 + 1 = 122 g mol^-1
Thus, 1000 mL of solution contains benzoic acid = 18.3g
250 mL of solution will contain benzoic acid
= \(\frac { 18.3 }{ 1000 }\) x 250 = 4.575 g

Question 13.
A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86°C, whereas pure ether boils at 35.60 °C. Determine the molecular mass of the solute (For ether K_b 2.02 K kg mol^-1)
Solution:
We have, mass of solute, W₂ = 8 g
Mass of solvent, W₁ = 100 g
Elevation of boiling point
∆T_b = 36.86 – 35.60 = 1.26⁰C
K_b = 2.02
Molecular mass of the solute

= 128.25g mol^-1

Question 14.
Ethylene glycol (molar mass = 62 g mol^-1) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4 of this substance in 100 g of water. Would it be advisable to keep this substance ¡n the car radiator during summer? (K_f for water = 1.86 k kg/mol^-1) (K_b for water = 0.512 K kg/mol^-1)
Solution:

Since water boils at 100°C, so a solution containing ethylene glycol will boil at 101.024 °C, SO it is advisable to keep this substance in car radiator during summer.

Question 15.
15.0 g of an unknown molecular material is dissolved in 450g of water. The resulting solution freezes at – 0.34°C. What is the molar mass of the material? K_f for water = 1.86 K kg mol^-1.
Solution:
W_(solute) = 1 5.0 g
W_(solvent) = 450 g
∆T_f = T°_f – T_f = 0 – ( – 0.34) = 0.34 °C
∆T_f = k_f m
0.34 = 1.86 x \(\frac { 15 }{ M }\) x \(\frac { 1000 }{ 450 }\)
M = \(\frac { 1.86 x 15 x 1000 }{ 0.34 x 450 }\) = 182.35 g mol^-1

Common Errors
Common Errors:

1. Students are writing solute, solvent, students get confused to write A (or) B, 1 (or)2.
2. Mole, mole fraction may be confused.
3. In writing osmosis definition Students get confused in mentioning concentration terms.
4. van’t Hoff factor I formula may be con fused.
5. Students may get contused rhen they write solute and solvent.
6. Mole and mole fraction may be confused by students.
7. Standard solutions must be known.
8. When students write Raoult’s law, they get confused with solute and solvent.
9. When they write the definition of osmosis, the conc. term may be little con fused.
10. van’t Hoff equation may be written wrongly.

Rectifications:

1. Always solvent is first so it is denoted as A (or) 1 solute is second so, it is denoted as B (or)2.
2. Mole = n ; Mole fraction x
3. Osmosis-movenient of solvent from low concentration to high concentration through a semipermeable membrane.
4. 
5. For e.g.. solid in liquid means solid is the solute and liquid is the solvent.
6. 
7. 1 N = 1 normal solution, 0.1 N Decinormal solution, 0.01 N = Centinormal solution
8. In Raoult’s law, “A” is always solvent and “B” is always solute.
9. In osmosis, always solvent rnoes through semi permeable membrane from low concentration to high concentration.
10. van’t Hoff factory = i
    

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Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Multiple Choice Questions

Question 1.
If K_b and K_f for a reversible reactions are 0.8 x 10^-5 and 1.6 x 10^-4 respectively, the value of the equilibrium constant is
(a) 20
(b) 0.2 x 10^-1
(c) 0.05
(d) None of these
Answer:
(a) 20
Solution:

Question 2.
At a given temperature and pressure, the equilibrium constant values for the equilibria

The relation between K₁ and K₂ is

Answer:

Solution:

Question 3.
The equilibrium constant for a reaction at room temperature is K₁ and that at 700 K is K₂. If K₁ > K₂ then
(a) The forward reaction is exothermic
(b) The forward reaction is endothermic
(c) The reaction does not attain equilibrium
(d) The reverse reaction is exothermic
Answer:
(a) The forward reaction is exothermic
Solution:
T₁ = 25 + 273 = 298 K, T₂ = 700 K

In this case, T₂ > T₁ and K₁ > K₂

ΔH° is – ve i.e., forward reaction is exothermic.

Question 4.
The formation of ammonia from N₂(g) and H₂(g) is a reversible reaction
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) + Heat
What is the effect of increase of temperature on this equilibrium reaction
(a) equilibrium is unaltered
(b) formation of ammonia is favoured
(c) equilibrium is shifted to the left
(d) reaction rate does not change
Answer:
(c) equilibrium is shifted to the left
Solution:
Increase in temperature, favours the endothermic reaction, given that formation of NH₃ is exothermic i.e., the reverse reaction is endothermic.
∴ Increase in temperature, shift the equilibrium to left

Question 5.
Solubility of carbon dioxide gas in cold water can be increased by ………….
(a) increase in pressure
(b) decrease in pressure
(c) increase in volume
(d) none of these
Answer:
(a) increase in pressure
Solution:

increase in pressure, favours the forward reaction.

Question 6.
Which one of the following is incorrect statement?
(a) for a system at equilibrium, Q is always less than the equilibrium constant
(b) equilibrium can be attained from either side of the reaction
(c) presence of catalyst affects both the forward reaction and reverse reaction to the same extent
(d) Equilibrium constant varied with temperature
Answer:
(a) for a system at equilibrium, Q is always less than the equilibrium constant
Solution:
Correct statement is, for a system at equilibrium, Q = K_eq

Question 7.
K₁ and K₂ are the equilibrium constants for the reactions respectively.

What is the equilibrium constant for the reaction NO₂(g) \(\rightleftharpoons\) \(\frac { 1 }{ 2 }\)N₂(g) + O₂(g)

Answer:

Solution:
Let equilibrium constant for the required reaction be K. Then,

Question 8.
In the equilibrium, 2A(g) \(\rightleftharpoons\) 2B(g) + C₂(g) the equilibrium concentrations of A, B and C, at 400 K are 1 x 10⁴ M, 2.0 x 10^-3 M, 1.5 x 10^-4 M respectively. The value of K_c. for the equilibrium at 400 K is
(a) 0.06
(b) 0.09
(c) 0.62
(d) 3 x 10^-2
Answer:
(a) 0.06
Solution:
[A] = 1 x 10^-4 M
[B] = 2 x 10^-3 M
[C] = 1.5 x 10^-4 M

Question 9.
An equilibrium constant of 3.2 x 10^-6 for a reaction means, the equilibrium is ………
(a) largely towards forward direction
(b) largely towards reverse direction
(c) never established
(d) none of these
Answer:
(b) largely towards reverse direction
Solution:

Question 10.
\(\frac { { K }_{ c } }{ { K }_{ p } }\) for the reaction, N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) is ………
(a) \(\frac { 1 }{ RT }\)
(b) \(g\sqrt { RT } \)
(c) RT
(d) (RT)²
Answer:
(d) (RT)²
Solution:
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
∆n_g = 2 – 4 = – 2
K_p = K_c (RT)^-2
\(\frac { { K }_{ c } }{ { K }_{ p } }\) = (RT)²

Question 11.
For the reaction, AB(g) \(\rightleftharpoons\) A(g) + B(g), at equilibrium, AB is 20% dissociated at a total pressure of R The equilibrium constant K is related to the total pressure by the expression ………..
(a) P = 24 K_p
(b) P = 8 K_p
(c) 24 p = K_p
(d) none of these
Answer:
(a) P =24 K_p
Solution:

Question 12.
In which of the following equilibrium, K and K are not equal?
(a) 2NO(g) \(\rightleftharpoons\) N₂(g) + O₂(g)
(b) SO₂(g) + NO₂(g) \(\rightleftharpoons\) SO₃(g) + NO(g)
(c) H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
(d) PCI₅(g) \(\rightleftharpoons\) PCl₃(g) + Cl₂(g)
Answer:
(d) PCI₅(g) \(\rightleftharpoons\) PCl₃(g) + Cl₂(g)
Solution:
For reactions given in options (a), (b) and (c) ∆n_g = 0
For option (d) ∆n_g = 2 – 1 = 1
∴ K_p = K_c (RT)

Question 13.
If x is the fraction of PCI₅ dissociated at equilibrium in the reaction,
PCl₅ \(\rightleftharpoons\) PCl₃ + Cl₂
then starting with 0.5 moIe of PCI₅ the total number of moles of reactants and products at equilibrium is
(a) 0.5 – x
(b) x + 0.5
(e) 2x + 0.5
(d) x ± 1
Answer:
(b) x + 0.5
Solution:

Total no. of moles at equilibrium = 0.5 – x + x + x = 0.5 + x

Question 14.
The values of K_p1 and K_p2 for the reactions X \(\rightleftharpoons\) Y + Z and A \(\rightleftharpoons\) 2B are in the ratio 9 : 1 if degree of dissociation and initial concentration of X and A be equal then total pressure at equilibrium P₁, and P₂ are in the ratio
(a) 36 : 1
(b) 1: 1
(c) 3 : 1
(d) 1: 9
Answer:
(a) 36: 1
Solution:

Question 15.
In the reaction, Fe(OH)₃(s) \(\rightleftharpoons\) Fe³⁺(aq) + 3OH^–(aq), if the concentration of OH^– ions is decreased by ¼ times, then the equilibrium concentration of Fe³⁺ will …………
(a) not changed
(b) also decreased by ¼ times
(c) increase by 4 times
(d) increase by 64 times
Answer:
(d) increase by 64 times
Solution:

[∵ Concentration of solids is constant)
When concentration of OH^– ions decreased by \(\frac { 1 }{ 4 }\) times, then

To maintain K_C as constant, concentration of Fe³⁺ will increase by 64 times.

Question 16.
Consider the reaction where K_p = 0.5 at a particular temperature
PCI₅(g) \(\rightleftharpoons\) PCI₃(g) + Cl₂(g) if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true?
(a) more PCI₃ will be produced
(b) more Cl₂ will be produced
(c) more PCI₅ will be produced
(d) none of these
Answer:
(c) more PCI₅ will be produced .
Solution:
K_p = 0.5

Q = \(\frac { 1 x 1 }{ 1 }\)
Q > K_p
∴Reverse reaction is favoured; i.e., more PCI₅ will be produced.

Question 17.
Equimolar concentrations of H₂ and I₂ are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of H₂ has reacted at equilibrium if rate constant for both forward and reverse reactions are equal
(a) 33%
(b) 66%
(c)(33)² 0/0
(d) 16.5%
Answer:
(a) 33%
Solution:

As degree of dissociation cannot be negative. Therefore, degree of dissociation
= \(\frac { a }{ 3 }\) x 100 = 33.33%

Question 18.
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 x 1 and the equilibrium constant is 50. The rate constant for the reverse reaction is ……….
(a) 11.5
(b) 5
(c) 2 x 10²
(d) 2 x 10³
Answer:
(b) 5
Solution:

Question 19.
Which of the following is not a general characteristic of equilibrium involving physical process
(a) Equilibrium is possible only in a closed system at a given temperature
(b) The opposing processes occur at the same rate and there is a dynamic but stable condition
(c) All the physical processes stop at equilibrium
(d) All measurable properties of the system remains constant
Answer:
(c) All the physical processes stop at equilibrium
Solution:
Correct statement – Physical processes occurs at the same rate at equilibrium.

Question 20.
For the fórmation of two moles of SO₃(g) from SO₂ and O₂, the equilibrium constant is K₁. The equilibrium constant for the dissociation of one mole of SO₃ into SO₂ and O₂ is …………

Answer:

Solution:
2SO₂ +O₂ \(\rightleftharpoons\) 2SO₃

Dissociation of 1 mole of 2SO₃

Question 21.
Match the equilibria with the corresponding conditions …………
(i) Liquid \(\rightleftharpoons\) Vapour – 1. Melting point
(ii) Solid \(\rightleftharpoons\) Liquid – 2. Saturated solution
(iii) Solid \(\rightleftharpoons\) Vapour – 3. Boiling point
(iv) Solute (s) \(\rightleftharpoons\) Solute (Solution) – 4. Sublimation point
5. Unsaturated solution

Answer:
(b) 3, 1, 4, 2

Question 22.
Consider the following reversible reaction at equilibrium, A + B \(\rightleftharpoons\) C, if the concentration of the reactants A and B are doubled, then the equilibrium constant will ………
(a) be doubled
(b) become one fourth
(c) be halved
(d) remain the same
Answer:
(d) remain the same
Solution:
A + B \(\rightleftharpoons\) C
K_C = \(\frac { [C] }{ [A] [B] }\)
If [A] and [B] are doubled, [C] increases 4 times to maintain K_C as constant.
∴ Equilibrium constant will remain the same.

Question 23.
[CO(H₂O)₆]²⁺ (aq) (pink) + 4C^– (aq) \(\rightleftharpoons\) [COCI₄]^2- (aq) (blue) + 6H₂O (1) In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in colour. On the basis of this information, which one
of the following is true?
(a) ∆H > 0 for the forward reaction
(b) ∆H = 0 for the reverse reaction
(c) ∆H < 0 for the forward reaction
(d) Sign of the ∆H cannot be predicted based on this information.
Answer: (a) ∆H > 0 for the forward reaction
Solution:
On cooling, reverse reaction predominates and the solution is pink in colour. Decrease in temperature, favours the reverse reaction i.e. reverse reaction is exothermic (∆H < 0)and for the forward reaction is endothermic (∆H > 0).

Question 24.
The equilibrium constants of the following reactions are:
N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃ ; K₁
N₂ + O₂ \(\rightleftharpoons\) 2NO ; K₂
H₂+ \(\frac { 1 }{ 2 }\)O₂ \(\rightleftharpoons\) H₂O ; K₃
The equilibrium constant (K) for the reaction;  will be

Answer:

Solution:

Question 25.
A 20 litre container at 400 K contains CO₂(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO₂ attains its maximum value will be.
Given that: SrCO₃(s) \(\rightleftharpoons\) SrO(s) + CO₂(g)
(a) 2 litre
(b) 5 litre
(c) 10 Litre
(d) 4 litre
Answer:
(b) 5 litre
Solution:
Given that K_p = 1.6 atm
V₁ = 20 L
V₂ = ?
T₁ = 400 K
T₂ = 400 K
K_p = P_co2
P_co2 = 1.6 atm
P₁ = 0.4 atm
P₂ = 1.6 atm

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Short Answer Questions

Question 26.
If there is no change in concentration, why is the equilibrium state considered dynamic?
Answer:
At chemical equilibrium the rate of two opposing reactions are equal and the concentration of. reactants and products do not change with time. This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.

Question 27.
For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain.
Answer:
In the chemical reaction, as the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches the equilibrium. So even at particular temperature, Q is not constant. Even once the equilibrium is achieved then change in concentration of reactants or products, pressure, volume will change the value of Q.

Question 28.
What is the relation between K_p and K_C. Give one example for which K_p is equal to K_C.
Answer:
The relation between K_p and K_c is K_p = K_C (RT)^∆n_g
K_p = equilibrium constant is terms of partial pressure.
K_c = equilibrium constant is terms of concentration.
R = gas constant
T = Temperature.
∆n_g = Difference between the sum of the number of moles of products and the sum of number of moles of reactants in the gas phase. When ∆n_g = 0
K_p = K_C(RT)⁰ = K_C i.e., K_p = K_C
Example: H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
∆n_g = 2 – 2 = 0
∴ K_p = K_C for the synthesis of HI.

Question 29.
For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is K_C is larger or smaller than K_p ?
Answer:
For a homogeneous reaction at equilibrium, number of moles of products (n_p) are greater than the number of moles of reactants (n_R) then ∆n_g = + ve
n_p > n_R
∆n_g = + ve
If ∆n_g is ve, K_p value is greater than K_C
K_p = K_C. (RT)^+ve
K_p > K_C
Example: PCl₅ \(\rightleftharpoons\) PCl₃(g) + Cl₂(g)
2 – 1 = 1
K_p = K_C (RT)¹
K_p > K_C

Question 30.
When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant (K), in which direction does the reaction proceed to reach equilibrium?
Answer:
When Q > K_C the reaction will proceed in the reverse direction, i.e, formation of reactants.

Question 31.
For the reaction: A₂(g) + B₂(g) \(\rightleftharpoons\) 2AB(g); ∆H is – ve.
Answer:
The following molecular scenes represent different reaction mixture
(A – dark grey, B – light grey)

1. Calculate the equilibrium constant K_p and K_C
2. For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
3. What is the effect of increase in pressure for the mixture at equilibrium?

Answer:
1.
At equilibrium

2.

3.  Since ∆n_g = 2 – 2 = 0, thus, pressure has no effect. So by increasing the pressure, equilibrium will not be affected.

Question 32.
State Le – Chatelier principle.
Answer:
It states that If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.

Question 33.
Consider the following reactions

1. H₂(g) + l₂(g) \(\rightleftharpoons\) 2HI(g)
2. CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₂(g)
3. S(s) + 3F₂(g) \(\rightleftharpoons\) SF₆(g)

In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.
Answer:
1.  H₂(g) + l₂(g) \(\rightleftharpoons\) 2HI(g)
In the above equilibrium reaction, volume of gaseous molecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

2.
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product CO₂, the pressure should be decreased or volume should be increased.

3.
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 34.
State law of mass action.
Answer:
The law states that “At any instant, the rate of chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant.”

Question 35.
Explain how will you predict the direction of an equilibrium reaction.
Answer:

1. A large value of K_C indicates that the reaction reaches equilibrium with high product yield.
2. A low value of K_C indicate that the reaction reaches equilibrium with low product formed.
3. In general, if the K is greater than I the reaction proceeds nearly to completion. If is less than 10^-3, the reaction rarely proceeds.
4. If K < 10^-3, reverse reaction is favoured. If K_c > 10³, forward reaction is favoured.

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Long Answer Questions

Question 36.
Derive a general expression for the equilibrium constant K_p and K_C for the reaction.
3H₂(g) + N₂(g) \(\rightleftharpoons\) 2NH₃(g)
Answer:
In the formation of ammonia, ‘a’ moles of Nitrogen and ‘b’ moles of hydrogen gas are allowed to react in a container of volume of ‘V’. Let ‘x’ moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia.
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)

Applying law of mass action

Total number of moles at equilibrium n = a – x + b -3x + 2x = a + b – 2x

Total number of moles at equilibrium
n = a – x + b – 3x + 2x = a + b – 2x

Question 37.
Write a balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression?
Answer:

Question 38.
What is the effect of added inert gas on the reaction at equilibrium at constant volume?
Answer:
When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles. of gases present in the container increases, that is, the total pressure of gases increases, the partial pressure of the reactants and the products are unchanged. Hence at constant volume, addition of inert gas has no effect on equilibrium.

Question 39.
Derive the relation between K_p and K_C.
Answer:
Consider a general reaction in which all reactants and products are ideal gases.
x A + y B \(\rightleftharpoons\)  lC+mD
The equilibrium constant K_C is

The ideal gas equation is
PV = nRT or P = \(\frac { n }{ V }\) RT
Since,
Active mass = molar concentration = \(\frac { n }{ V }\)
P = Active mass x RT
Based on the above expression, the partial pressure of the reacants and products can be expressed as

On substitution in equation (2),

By comparing equation (1) and (4), we get
K_p = K_C (RT) _∆ng
where ∆n_g is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
(i) If ∆n_g = 0, K_b = K_C(RT)⁰
K_p = K_C
Example: H₂(g) + I₂ \(\rightleftharpoons\) 2HI(g)

(ii) where
∆n_g = +Ve
K_p = K_C (RT)^+ve
K_p = K_C
Example: 2NH₃(g) N₂(g) + 3H₂(g)

(iii) When
∆n_g = -Ve
K_p = K_C (RT)^-ve
K_p < K_C
Example: 2SO₂(g) +O₂(g) \(\rightleftharpoons\) 2SO₃(g)

Question 40.
One mole of PCl₅ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.
Answer:
Given that

Question 41.
For the reaction: SrCO₂(s) \(\rightleftharpoons\) SrO (s) + CO₂(g), the value of equilibrium constant K_p = 2.2 x 10^-4 at 1002K. Calculate K_C for the reaction.
Answer:
For the reaction, SrCO₂(s) \(\rightleftharpoons\) SrO (s) + CO₂(g)
∆n_g = 1 – 0 = 1
K_p = K_p (RT)
2.2 x 10^-4 = K_C (0.0821)(1002)

Question 42.
To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate K_p and K_C for this reaction.
Answer:

Question 43.
Oxidation of nitrogen monoxide was studied at 200°C with initial pressures of 1 atm NO and 1 atm of O₂. At equilibrium partial pressure of oxygen is found to be 0.52 atm. Calculate K_p value.
Answer:
2NO(g) + O₂(g) \(\rightleftharpoons\) 2NO₂(g)

Question 44.
1 mol of CH₄, 1 mole of CS₂ and 2 mol of H₂S are 2 moI of H₂ are mixed in a 500 mL flask. The equilibrium constant for the reaction K_C = 4 x 10^-2 mol² lit^-2. In which direction will the reaction proceed to reach equilibrium?
Answer:

∴ The reaction will proceed in the reverse direction to reach the equilibrium.

Question 45.
At particular temperature K_C = 4 x 10^-2 for the reaction
H₂S(g) \(\rightleftharpoons\) 2H₂(g) + \(\frac { 1 }{ 2 }\) S₂(g). Calculate K_C for each of the following reaction
(i). H₂S(g) \(\rightleftharpoons\) 2H₂(g) + S₂(g)
(ii). 3H₂S(g) \(\rightleftharpoons\) 3H₂(g) + \(\frac { 3 }{ 2 }\) S₂(g)
Answer:
K_C = 4 x 10^-2 for the reaction

For the reaction 3H₂S(g) \(\rightleftharpoons\) 3H₂(g) + \(\frac { 3 }{ 2 }\) S₂(g)

Question 46.
28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17g NH₃ was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.
Answer:
Given

[NH₃] = \((\frac { 17 }{ 17 })\) = 1 mol = 2x
x = 0.5 mol
At equilibrium, [N₂] = 1 – x = 0.5 mol
[H₂] = 3 – 3x = 3 – 3 (0.5) = 1.5 mol
Weight of N₂ (no. of moles of N₂) x molar mass of N₂ = 0.5 x 28 = 14g
Weight of H₂ = (no. of moles of H₂) x molar mass of H₂= 1.5 x 2 = 3g

Question 47.
The equilibrium for the dissociation of XY₂ is given as,
2XY₂ (g) \(\rightleftharpoons\) 2XY (g) + Y₂(g).
If the degree of dissociation x is so small compared to one. Show that 2 K_p = Px³ where P is the total pressure and K is the dissociation equilibrium constant of XY₂.
Answer:
2XY₂ (g) \(\rightleftharpoons\) 2XY (g) + Y₂(g)


Question 48.
A sealed container was filled with I mol of A, (g), I mol B, (g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 1 for the reaction:
A₂(g) + B₂(g) \(\rightleftharpoons\) 2AB(g)
Answer:
A₂(g) + B₂(g) \(\rightleftharpoons\) 2AB(g)


[B₂]_eq = 1 – x = 1 – 0.33 = 0.667
[AB]₂ = 2x = 2 x 0.33 = 0.66
Question 49.
Deduce the Vant Hoff equation.
Answer:
This equation gives the quantitative temperature dependance of equilibrium constant K. The relation between standard free energy change ∆G° and equilibrium constant is
∆G° = – RT ln K  …………(1)
We know that,
∆G° = ∆H° – T∆S°        ………….(2)
Substituting (1) in equation (2)
– RT lnK = ∆H° – T∆S°

Rearranging, ln K = \(\frac { – ∆H° }{ RT }\) + \(\frac { ∆S° }{ R }\)   ………….(4)
Differentiating equation (3) with respect to temperature
d(lnK) ∆H° \(\frac { d(lnK) }{ dT }\) = \(\frac { { \triangle H }^{ 0 } }{ { RT }^{ 2 } }\)
Equation (4) is known as differential form of van’t Hoff equation. On integrating the equation 4, between T₁ and T₂ with their respective equilibrium constants and K₂

Equation is known as integrated form of van’t Hoff equation.

Question 50.
The equilibrium constant K for the reaction
N₂(g) + 3H₂(g) = 2NH₃(g) is 8.19 x 10²
at 298 K and 4.6 x 10^-1 at 498 K. Calculate ∆H° for the reaction.
Answer:

Question 51.
The partial pressure of carbon dioxide in the reaction CaCO₃(s) \(\rightleftharpoons\) CaO (s) + CO₂(g) is 1.017 x 10^-3 atm at 500° C. Calculate K_p, at 600° C for the reaction. ∆H for the reaction is 181 kJ mol^-1 and does not change in the given range of temperature.
Answer:

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium In – Text problems Solved

Question 1.
Consider the following reaction
Fe³⁺(aq) + SCN^–(aq) [Fe(SCN)]²⁺(aq)
A solution is made with initial Fe³⁺, SCN^– concentration of 1 x 10^-3 M and 8 x 10^-4M respectively. At equilibrium [Fe(SCN)]²⁺ concentration is 2 x 10^-4M. Calculate the value of equilibrium constant.
Answer:

Question 2.
The atmospheric oxidation of NO
2NO(g) + O₂(g) \(\rightleftharpoons\) 2NO₂(g)
was studied with initial pressure of 1 atm of NO and I atm of O₂. At equilibrium, partial pressure of oxygen is 0.52 atm. Calculate K_p of the reaction.
Answer:

As,
1 – x = 0.52
x = 0. 48
= At equilibrium,
P_NO = 1 – 2x = 1 – 2(0.48) = 0.04
P_NO2 = 2x = 2(0.48) = 0.96

Question 3.
The following water gas shift reaction is an important industrial process for the production of hydrogen gas.
CO(g) + H₂O(g) \(\rightleftharpoons\) CO₂(g) + H₂(g) At a given temperature K_p = 2.7.  If 0.13 moI of CO, 0.56 moI of water, 0.78 mol of CO₂ and 0.28 mol of H₂ are introduced into a 2L flask, find out in which direction must the reaction proceed to reach equilibrium.
Answer:
CO(g) + H₂O(g) \(\rightleftharpoons\) CO₂(g) + H₂(g)
Given
K_p = 2.7
[CO] = 0.13 mol, [H₂O] = 0.56 mol
[CO₂] = 0.78 mol; [H₂] = 0.28 mol
V = 2L
K_p = K_C (RT)^Δn_g
2.7 = K_C(RT)°
K_C = 2.7

Q = 3
Q > K_C Hence, the reaction proceed in the reverse direction.

Question 4.
1 moI of PCl₅, kept in a closed container of volume 1 dm³ and was allowed to attain equilibrium at 423 K. Calculate the equilibrium composition of reaction mixture. (The K_C value for PCl₅ dissociation at 423 K is 2)
Answer:
PCl₅ \(\rightleftharpoons\) PCL₃ + Cl₂
Given that [PCl₅]_initial = 1 mol
V = 1 dm³
K_C = 2

∴ Equilibrium concentration of
[PCl₅]_eq = \(\frac { 1-x }{ 1 }\) = 1 – 0.732 = 0.268 M
[PCl₃]_eq = \(\frac { x }{ 1 }\) = \(\frac { 0.732 }{ 1 }\) = 0.732
[Cl₂]_eq = \(\frac { x }{ 1 }\) = \(\frac { 0.732 }{ 1 }\) = 0.732

Question 5.
The equilibrium constant for the following reaction is 0.15 at 298 K and 1 atm pressure.
Answer:
N2O₄(g) \(\rightleftharpoons\) 2NO₂(g)
T₁ = 298 K
K_p1 = 0.15
T₂ = 100⁰C = 100 + 273 = 373 K
K_p2 = ?

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium In – Text problems Solved

Question 1.
One mole of H₂ and one mole of I₂ are allowed to attain equilibrium in 1 lit container. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.
Answer:
Given data: [H₂] = I mole, [I₂] = 1 mole
At equilibrium, [HI] = 0.4 mole, K_C = ?
H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)

Question 2.
The equilibrium concentrations of NH₃, N₂ and H₂ are 1.8 x 10² M, 1.2 x 10^-2 M and 3 x 10^-2 M respectively. Calculate the equilibrium constant for the formation of NH₃ from N₂ and H₂. [Hint: M = mol lit^-1]
Answer:
Given data:
[NH₃] 1.8 x 10^-2 M, [N₂] = 1.2 x 10^-2M, [H₂] 3 x 10^-2 M, K_C = ?

Question 3.
The equilibrium constant at 298 K for a reaction is 100.
A+ B \(\rightleftharpoons\) C + D
If the initial concentration of all the four species is 1 M, the equilibrium concentration of D (in mol lit^-1) will be
Answer:
Given data: [A] = [B] = [C] = [D] = 1 M, K_C = 100, [D]_eq = ?
Let x be the number of moles reactants reacted

Question 4.
For an equilibrium reaction K_p = 0.0260 at 25° C and ?H = 32.4 kJ mor^-1. Calculate K_p at 37° C.
Answer:
T₁ = 25 + 273 = 298 K
T₂ = 37 + 273 = 310 K
ΔH = 32.4 kJ mor^-1 = 32400 J mol^-1
R = 8.314 JK^-1 mol^-1
K_p1 = 0.0260
K_p2 = ?

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Additional Questions 

I. Choose the correct answer.

Question 1.
Which of the following represents physical equilibrium?
(a) PCl₅(g) \(\rightleftharpoons\) PCl₃(g) + Cl₂(g)
(b) H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
(c) H₂O(l) \(\rightleftharpoons\) H₂O(g)
(d) N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
Answer:
(c) H₂O(l) \(\rightleftharpoons\) H₂O(g)
Solution:
Physical states arc in equilibrium i.e., liquid – vapour equilibrium.

Question 2.
Which one of the following is an example of chemical equilibrium?
(a) 2NO(g) + O₂(g) \(\rightleftharpoons\) 2NO₂(g)
(b) I₂(s) \(\rightleftharpoons\) I₂(g)
(c) H₂O(s) \(\rightleftharpoons\) H₂O(1)
(d) NH₂CI(s) \(\rightleftharpoons\) NH₄CI(g)
Answer:
(a) 2NO(g) + O₂(g) \(\rightleftharpoons\) 2NO₂(g)
Solution:
All the other three are physical equilibrium. Only (a) is chemical equilibrium.

Question 3.
Which one of the following does not undergo sublimation?
(a) Iodine
(b) water
(c) Camphor
(d) Ammonium chloride
Answer:
(b) Water

Question 4.
At chemical equilibrium,
(a) rate of forward reaction = rate of backward reaction
(b) rate of forward reaction > rate of backward reaction
(c) rate of forward reaction < rate of backward reaction
(d) rate of forward reaction = rate of backward reaction
Answer:
(a) rate of forward reaction = rate of backward reaction

Question 5.
Which of the following is an example of homogeneous equilibrium?
(a) H₂O(1) \(\rightleftharpoons\) H₂O(g)
(b) CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO(g)
(c) H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
(d) 2CO(g) \(\rightleftharpoons\) CO₂(g) + C(s)
Answer:
(c) H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
Solution:
Here all reactants and products are in same phase i.e., gaseous phase.

Question 6.
Which of the following is an example of heterogeneous equilibrium?
(a) Synthesis of HI
(b) Dissociation of PCI₅
(c) Acid hydrolysis of ester
(d) Decomposition of limestone
Answer:
(d) Decomposition of limestone
Solution:
CaCO₃(s) → CaO(s) + CO₂(g)
Here CO₂ is in gaseous state while CaCO₃ and CaO are in solid state.

Question 7.
Statement I: In dissociation of PCI₅ to PCI₃ and CI₂, K_p > K_C
Statement II: In dissociation of PCI₅, Δn_g = – ve and so K_p > K_C.
(a) Statement I & II are correct and statement II is the correct explanation of statement I.
(b) Statement I & II are correct but statement II is not the correct explanation of statement I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(c) Statement I is correct but statement II is wrong.
Solution:
Δn_g = 2 – 1 = 1 = +ve

Question 8.
In the reaction, 2NH₃(g) \(\rightleftharpoons\) N₂(g) + 3H₂(g)
(a) K_p = K_C
(b) K_p < K_C
(c) K_p > K_C
(d) K_p = \(\frac { 1 }{ { K }_{ C } }\)
Answer:
(c) K_p > K_C
Solution:
K_p = K_C (RT)^Δn_g and Δn_g = 4 – 2 = 2
∴ K_p = K_C (RT)² = K_p > K_C

Question 9.
In which of the following reaction, K_p is equal to K_C ?
(a) N₂(g) ± O₂(g) \(\rightleftharpoons\) 2NO(g)
(b) 2NH₃(g) \(\rightleftharpoons\) N₂(g) + 3H₂(g)
(c) 2H₂(g) + O₂(g) \(\rightleftharpoons\) 2H₂O(g)
(d) PCI₅(g) \(\rightleftharpoons\) PCI₃(g) + CI₂
Answer:
(a) N₂(g) ± O₂(g) \(\rightleftharpoons\) 2NO(g)
Solution:
K_C (RT)^Δn_g when Δn_g = 0 then K = K for option (a), Δn_g = 2 – 2 = 0

Question 10.
In the equilibrium reaction CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₂(g) whose concentration remains constant at a given temperature?
(a) CaO
(b) CO₂
(e) CaCO₃
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)
Solution:
Concentration of solids remains constant at a particular temperature.

Question 11.
Consider the following equilibrium reaction and relate their equilibrium constants
(i) N₂ + O₂ \(\rightleftharpoons\) 2NO; K₁
(ii) 2NO + O₂ \(\rightleftharpoons\) 2NO₂; K₂
(iii) N₂ + 2O₂ \(\rightleftharpoons\) 2NO₂; K₃
(a) K₃ = K₂ = K₁
(b) K₁ x K₃ = K₂
(c) K₁ x K₂ = K₃
(d) \(\frac { { K }_{ 1 } }{ { K }_{ 2 } }\) = K₃
Answer:
(c) K₁ x K₂ = K₃
Solution:

Question 12.
Statement I: A pure solid in an equilibrium reaction has the same concentration at a given temperature.
Statement II: The solid does not expand to fill its container and it has same number of moles of its volume.
(a) Statement I and II are correct and statement II is the correct explanation of statement of I.
(b) Statement I and II are correct but II is not the correct explanation of!.
(c) Statement I and II are not correct.
(d) Statement I is wrong but 11 is correct.
Answer:
(a) Statement I and II are correct and statement II is the correct explanation of statement of I.

Question 13.
Find the Q value of the reaction H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g) at an instant where concentration of H₂, I₂ and HI are found to be 0.2 moI L^-1, 0.2 mol L^-1, and 0.6 moI L^-1 respectively.
(a) 48
(b) 9
(c) 0.9
(d) 90
Answer:
(b) 9
Solution:

Question 14.
For the reaction N₂O₄(g) \(\rightleftharpoons\) 2NO₂(g) K_C = 0.21 at 373 K. The concentrations of N₂O₄ and NO₂ are found to be 0.125 mol dm^-3 and 0.5 mol dm^-3 respectively at a given temperature. Predict the direction of the reaction.
(a) At equilibrium
(b) reverse direction
(c) forward direction
(d) Both reverse and forward direction
Answer:
(b) reverse direction
Solution:

K_C = 0.21
Q = 2
Q > K_C
Hence the reaction will proced in the reverse direction

Question 15.
Which of the following does not alter the equilibrium?
(a) catalyst
(b) concentration
(c) temperature
(d) pressure
Answer:
(a) catalyst

Question 16.
Statement I. In Haber’s process, NH₃ is liquefied and removed.
Statement II. In manufacture of NH₃, liquefied and removal of NH₃, keeps the reaction moving in forward direction.
(a) Statement I and II are correct and II is the correct explanation of I.
(b) Statement I and II are correct but II is not the correct explanation of I.
(c) Statement I is wrong but statement II is correct.
(d) Statement I is correct but statement II is wrong.
Answer:
(a) Statement I and II are correct and II is the correct explanation of I.
Solution:
Removal of NH₃ will decrease its concentration which favours the production of NH₃ according to the Le – Chatelier’s principle.

Question 17.
In which of the following reaction, pressure has no effect?
(a) N₂ + 3N₂ \(\rightleftharpoons\) 2NH₃(g)
(b) 2SO₂(g) + O₂(g) \(\rightleftharpoons\) 2SO₃(g)
(c) N₂O₄(g) \(\rightleftharpoons\) 2NO₂(g)
(d) H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
Answer:
(d) H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
Solution:
In the reaction H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g) the volumes are equal on both sides and so pressure has no effect.

Question 18.
Among the following reactions which one has K_p = K_C
(a) N₂O₄ \(\rightleftharpoons\) 2NO(g)
(b) 2SO₂(g) + O₂(g) \(\rightleftharpoons\) 2SO₃(g)
(c) N₂(g) + O₂(g) \(\rightleftharpoons\) 2NO(g)
(d) N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
Answer:
(c) N₂(g) + O₂(g) \(\rightleftharpoons\) 2NO(g)
Solution:
K_p = K_C. RT^Δn_g
In equation N₂(g) + O₂(g) \(\rightleftharpoons\) 2NO(g)
Δn_g = 0
K_p = K_C. RT°
K_p = K_C

Question 19.
Statement I. Addition of an inert gas at constant volume has no effect on equilibrium.
Statement II. When an inert gas is added, the total number of moles of gases present in the container increases and total pressure also increases, the partial pressure of the products and reactants are unchanged.
(a) Statement I and II are correct but statement II is not the correct explanation of I.
(b) Statement I and II are correct and statement II is the correct explanation of 1.
(c) Statement I is correct but statement II is not correct.
(d) Statement I is wrong but statement II is correct.
Answer:
(b) Statement I and II are correct and statement II is the correct explanation of I.

Question 20.
Which one of the following equation is not correct?
(a) ΔG° = – RTInK
(b) ΔG° = ΔH° – TΔS°
(c) – RTInK = ΔH° – TΔS°
(d) In k = \(\frac { ΔH° }{ T }\) – \(\frac { ΔS° }{ R }\)
Answer:
(d) In k = \(\frac { ΔH° }{ T }\) – \(\frac { ΔS° }{ R }\)

Question 21.
The equilibrium expression, K_C = [CO₂] represents the reaction.
(a) C(s) + O₂(g) \(\rightleftharpoons\) CO₂(g)
(b) CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₂(g)
(c) 2CO(g) + O₂(g) \(\rightleftharpoons\) 2CO₂(g)
(d) CaO(s) + CO₂(g) \(\rightleftharpoons\) CaCO₃(s)
Answer:
(b) CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₂(g)

Question 22.
Hydrogen molecule (H₂) can be dissociated into hydrogen atoms (H). Which one of the following changes will not increase the number of atoms present at equilibrium?
(a) adding H atoms
(b) increasing the temperature
(c) increasing the total pressure
(d) increasing the volume of the container
Answer:
(d) increasing the total pressure container
Solution:
It favours backward reaction i.e., formation of H₂ molecule.

Question 23.
What is the expression for K_eq for the reaction, 2N₂O(g) + O₂(g) \(\rightleftharpoons\) 4NO(g)?

Answer:

Question 24.
What is the correct expression for the representation of the solubility product constant of Ag₂CrO₄ ?
(a) [Ag⁺]² [CrO₄^2-]
(b) [2Ag⁺] [CrO₄^2-]
(c) [Ag⁺] [CrO₄^2-]
(d) [2Ag⁺]² [CrO₄^2-]
Answer:
(a) [Ag⁺]² [CrO₄^2-]

Question 25.
H₂ + S \(\rightleftharpoons\) H₂S + energy. In this reversible reaction, select the factor which will shift the equilibrium to the right.
(a) adding heat
(b) adding H₂S
(c) blocking hydrogen gas reaction
(d) removing hydrogen suiphide gas
Answer:
(a) removing hydrogen suiphide gas

Question 26.
What effect does a catalyst have on the equilibrium position of a reaction?
(a) a catalyst favours the formation of products
(b) a catalyst favours the formation of reactants
(c) a catalyst does not change the equilibrium position of a reaction
(d) a catalyst may favour reactants or product formation, depending upon the direction in which the reaction is written.
Answer:
(c) a catalyst does not change the equilibrium position of a reaction

Question 27.
A chemist dissolves an excess of BaSO₄ in pure water at 25°C if its K_sp = 1 x 10^-10. What is the concentration of barium in the water?
(a) 10^-4 M
(b) 10^-5 M
(c) 10^-15 M
(a) 10^-6 M
Answer:
(c) 10^-15 M
Solution:
K_sp = [Ba²⁺] [SO₄^2-]
1 x 10^-10 = (x) (x)
10^-5 = x

Question 28.
If in a mixture where Q = K, then what happens?
(a) the reaction shift towards products
(b) the reaction shift towards reactants
(c) nothing appears to happen, but forward and reverse reactions are continuing at the same rate
(d) nothing happens
Answer:
(c) nothing appears to happen, but forward and reverse reactions are continuing at the same rate

Question 29.
If dissociation for reaction PCI₅ \(\rightleftharpoons\) PCI₃ + Cl₂ is 20% at 1 atm pressure. Calculate the value of K_C.
(a) 0.04
(b) 0.05
(c) 0.07
(d) 0.06
Answer:
(d) 0.05
Solution:

Question 30.
What would be the value of Δn_g for the reaction NH₄CI(s) \(\rightleftharpoons\) NH₂(g) + HCI(g)?
(a) 1
(b) 0.5
(c) 2
(d) 1.5
Answer:
(c) 2
Solution:
Δn_g = n_p – n_r = 2 – 0 = 2

Question 31.
Which of the following is not a general characteristic of equilibria involving physical processes?
(a) Equilibrium is possible only in a close system at a given temperature
(b) All measurable properties of the system remain constant
(c) All the physical processes stop at equilibrium
(d) The opposing processes occur at the same rate and there is dynamic but stable condition
Answer:
(c) All the physical processes stop at equilibrium

Question 32.
At 500K, equilibrium constant K_C for the following reaction is 5, \(\frac { 1 }{ 2 }\)H₂(g) + \(\frac { 1 }{ 2 }\)I₂(g) \(\rightleftharpoons\) HI(g) what would be the equilibrium constant K_C. for the reaction 2HI(g) \(\rightleftharpoons\) H₂(g) + I₂(g)
(a) 0.44
(b) 0.04
(c) 25
(d) 2.5
Answer:
(b) 0.04
Solution:

Question 33.
For the reaction 2NO₂(g) \(\rightleftharpoons\) 2NO(g) + O₂ (g), K_C = 1.8 x 10^-6 at 185⁰C. At the same temperature the value of K_C for the reaction. NO(g) + \(\frac { 1 }{ 2 }\)O₂ \(\rightleftharpoons\) NO₂(g) is ……….
(a) 0.9 x 10⁶
(b) 7.5 x 10²
(c) 1.95 x 10^-3
(d) 1.95 x 10³
Answer:
(b) 7.5 x 10²
Solution:
The reaction is reversed and halved.

Question 34.
Which of the following reaction will be favoured at low pressure?
(a) N₂ + O₂ \(\rightleftharpoons\) 2NO
(b) H₂ ± I₂ \(\rightleftharpoons\) 2HI
(c) PCI₅ \(\rightleftharpoons\) PCI₃ + Cl₂
(d) N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃
Answer:
(c) PCI₅ \(\rightleftharpoons\) PCI₃ + Cl₂

Question 35.
Consider the reaction CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₂(g) is a closed container at equilibrium. What would be the effect of addition of CaCO₃ on the equilibrium?
(a) increases
(b) remains unaffected
(c) decreases
(d) unpredictable
Answer:
(b) remains unaffected

Question 36.
For the reaction PCI₅(g) \(\rightleftharpoons\) PCI₃(g) + Cl₂(g) the forward reaction at constant temperature is favoured by …………………
(a) introducing an inert gas at constant volume
(b) introducing PCl₃(g) at constant volume
(c) introducing PCl₅(g) at constant volume
(d) introducing CI₂(g) at constant volume
Answer:
(c) introducing PCI₅(g) at constant volume

Question 37.
The equilibrium of the reaction N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) will shift to product side when….
(a) K_p > 1
(b) Q < K_p
(c) Q = K_p
(d) Q = 2K_p
Answer:
(b) Q < K_p

Question 38.
NO₂ is involved in the formation of smog and acid rain. A reaction that is important in the formation of NO₂ is O₃(g) + NO(g) \(\rightleftharpoons\) O₂(g) + NO₂(g) K_C = 6.0 x 10³⁴. If the air over a section of New Delhi contained 1.0 x 10^-6 M of O₃, 1.0 x 10^-5 M of NO, 2.5 x 10^-4 M of NO₂ and 8.2 x 10^-3 of O₂, what can we conclude?
(a) there will be a tendency to form more NO and O₂
(b) there will be a tendency to form more NO₂ and O₂
(c) there will be a tendency to form more NO₂ and O₃
(d) there will no tendency for change because the reaction is at equilibrium
Answer:
(b) there will be a tendency to form more NO₂ and O₂
Solution:

As Q < K_C the reaction will hive a tendency to move forward.

Question 39.
Haemoglobin (Hb) forms bond with oxygen and given oxyhaemoglobin (HbO₂). This process is partially regulated by the concentration of H₃O⁺ and dissolved CO₂ in blood as HbO₂ +H₃O⁺ + CO₂ \(\rightleftharpoons\) H⁺ – Hb – CO₂ + O₂ + H₂O. If there is production of lactic acid and CO₂ during a muscular exercise, then
(a) more HbO₂ is formed
(b) more O₂ is released
(c) CO₂ is released
(d) both (b) and (c)
Answer:
(b) more O₂ is released

Question 40.
In the reaction N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃ + x k Cal, one mole of N₂ reacts with. 3 moles of H₂ at equilibrium. Then the value of a (degree of dissociation) is approximately ……………… P is the pressure at equilibrium

Answer:

Solution:

II. Match the following.

Question 1.

Answer:
(a) 2, 1, 4, 3

Question 2.

Answer:
(b) 2, 3, 4, 1

Question 3.

Answer:
(a) 4, 1, 2, 3

Question 4.

Answer:
(a) 3, 4, 1, 2

Question 5.

Answer:
(b) 2, 3, 4, 1

III. Fill in the blanks.

Question 1.
Transport of oxygen by Hemoglobin in our body is ………… a reaction.
Answer:
reversible

Question 2.
The temperature at which the solid and liquid phases of a substance are at equilibrium is called …………
Answer:
Freezing point

Question 3.
Thejemperature at which the liquid and vapour phases are at equilibrium is called …………
Answer:
Condensation point

Question 4.
………… law is used to explain gas-solution equilibrium processes.
Answer:
Henry’ law

Question 5.
In the reaction 2H₂(g) + O₂(g) \(\rightleftharpoons\) 2H₂O(g), the K_p value is equal to ………
Answer:
< K_C
Solution:
Δn_g = 2 – 3 = -1

Qustion 6.
The expression of K for the reaction CO₂(g) + H₂O(l) \(\rightleftharpoons\) H⁺(aq) + HCO₃^–(aq) is equal to …………..
Answer:

Question 7.
The expression of K for the reversible reaction 2CO(g) \(\rightleftharpoons\) CO₂(g) + C(s)
Answer:

Question 8.
The Ang value for the reaction 2NO(g) + O₂(g) \(\rightleftharpoons\) 2NO₂(g) is …………….
Answer:
– 1
Solution:
Δn_g = n_p – n_r = 2 – 3 = -1

Question 9.
The correct differential form of van’t Hoff equation is ………….
Answer:

Question 10.
For the reaction H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g), the equilibrium constant K is
Answer:

Question 11.
PCI₅ is kept in a closed container at a temperature of 250K the equilibrium concentrations of PCI₅, PCl₃ and Cl₂ are 0.045 moles L^-1, 0.096 moles L^-1, 0.096 moles L^-1 respectively. The value of equilibrium constant for the reaction PCI₅(g) \(\rightleftharpoons\) PCl₃(g) + Cl₂(g) will be ……………………
Answer:
0.205
Solution:

Question 12.
Equilibrium constant changes with ………………..
Answer:
Both temperature and pressure

Question 13.
For the reaction 2HI(g) \(\rightleftharpoons\) H₂(g) + I₂(g) at 720 K, the equilibrium constant value is 50. The equilibrium constant for the reaction H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g) at the same temperature will be ………..
Answer:
0.02
Solution:
Forward reaction equilibrium constant K₁ = 50
Reverse reaction equilibrium constant K₂ = ?
K₂ = \(\frac { 1 }{ { K }_{ 1 } }\) = \(\frac { 1 }{ 50 }\) = 0.02

Question 14.
If equilibrium constant for the reaction N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) at 298 K is 2.54, the value of equilibrium constant for the reaction \(\frac { 1 }{ 2 }\)N₂ + \(\frac { 3 }{ 2 }\)H₂ \(\rightleftharpoons\) NH₃ will be ………….
Answer:
1.59
Solution:

Question 15.
The chemical system at equilibrium is not affected by addition of ……………..
Answer:
catalyst

Question 16.
A catalyst will increase the rate of a chemical reaction by lowering the ……………
Answer:
activation energy

Question 17.
In a closed system, A(s) \(\rightleftharpoons\) 3B(g) + 3C(g) If partial pressure of C is doubled, then partial pressure of B will be ………….. times the original value.
Answer:
\(\frac { 1 }{ 2\sqrt { 2 } }\)

Question 18.
Consider the following gaseous equilibria with equilibrium constants K₁ and K₂ respectively
SO₂ + \(\frac { 1 }{ 2 }\)O₂(g) \(\rightleftharpoons\) SO₃(g) – K₁
2SO₃(g) \(\rightleftharpoons\) 2SO₂(g) + O₂(g) – K₂
The equilibrium constants are related as ………….
Answer:

Solution:

Question 19.
K₂ for the following reaction at 700 K is 1.3 x 10^-3 atm^-1. The K at same temperature for the reaction 2SO₂(g) + O₂(g) \(\rightleftharpoons\) 2SO₃(g) will be ………….
Answer:
7.4 x 10^-2
Solution:

Question 20.
For the reaction PCI₂₃(g) + CI₂(g) \(\rightleftharpoons\) PCI₅(g) at 250°C, the value of K_C is 26 then the value of K_p on the same temperature will be ……………
Answer:
0.61
Solution:

Question 21.
In the reaction N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g), the value of the equilibrium constant depends on ………………
Answer:
the temperature

Question 22.
K₁ and K₂ are velocity constant o forward and backward reactions. The equilibrium constant K_C of the reaction is …………
Answer:
\(\frac { { K }_{ 1 } }{ { K }_{ 2 } }\)

Question 23.
The equilibrium constant of the reaction 3C₂H₂ \(\rightleftharpoons\) C₆H₆ is 4.0 at a temperature of T K. If the equilibrium concentration of C₂H₂ is 0.5 molesL^-1, the concentration of C₆H₆ is …………..
Answer:

Question 24.
In an equilibrium reaction for which ΔG⁰ = 0, the equilibrium constant K should be = ……………
Answer:
1

Question 25.
The equilibrium constant for the reaction 2SO₂(g) + O₂(g) \(\rightleftharpoons\) 2SO₃(g) is 5. If the equilibrium constant mixture contains equal moles of SO₃ and SO₂, the equilibrium partial pressure of O₂,
gas is ……………..
Answer:
0.2 atm
Solution:

Question 26.
In the reaction NH₄CI(s) \(\rightleftharpoons\) NH₃(g) + HCI(g) the value Offlg is Δn_{g ………………….}
Answer:
2
Solution:
Δn_g = n_p(g) – n_r(g) = 2 – 0 = 2

IV. Choose the odd one out.

Question 1.
(a) see – saw
(b) tug – of – war
(c) sublimation of camphor
(d) Acid hydrolysis of an ester
Answer:
(d) Acid hydrolysis of an ester Acid hydrolysis of an ester is a chemical equilibrium.

Whereas a, b, c are examples of physical equilibrium.

Question 2.
(a) Synthesis of hydrogen iodide
(b) Decomposition of calcium carbonate
(c) Sublimation of iodine
(d) Dissociation of PCl₅
Answer:
(c) Sublimation of iodine Sublimation of iodine I_2(s) \(\rightleftharpoons\) I_2(g) is an example of physical equilibrium, whereas a, b and dare examples of chemical equilibrium.

Question 3.
(a) Synthesis of HI
(b) Dissociation of PCl₅
(e) Synthesis of NH₃
(d) Decomposition of CaCO₃
Answer:
(a) Decomposition of CaCO₃
Decomposition of CaCO₃ is an example of heterogeneous equilibrium.
CaCO_3(s) \(\rightleftharpoons\) CaO_(s) + CO_2(g)
Where as a, b, c are examples of homogeneous equilibrium.

Question 4.

Answer:

Synthesis of SO₃ is an example of homogeneous equilibrium whereas the others a, b and d are heterogeneous equilibrium.

Question 5.

Answer:

(b) is an example of heterogeneous equilibrium whereas the others a, c and d are homogeneous equilibrium.

V. Choose the correct pair.

Question 1.
(a) Q = K_C : Reaction is in equilibrium state
(b) Q < K_C : Reaction proceed in reverse direction
(c) Q > K_C : Reaction proceed in ftrward direction
(d) Q = K_C : Reaction proceed in both directions
Answer:
(a) Q = K_C : Reaction is in equilibrium state

Question 2.

Answer:

Question 3.
(a) K_p = K_C : Synthesis of HI
(b) K_p > K_C : Dissociation of PCl₅
(c) K_p < K_C : Synthesis of SO₃
(d) K_p = K_C : Synthesis of HI
Answer:
(a) K_p = K_C : Synthesis of HI
H_2(g) + I_2(g) \(\rightleftharpoons\) 2HI_(g) : Δn_g = 2 – 2 = 0
ΔK_p = K_C

VI. Choose the incorrect pair.

Question 1.
(a) Acid hydrolysis of an ester – Homogeneous equilibrium
(b) Synthesis of Ammonia – Homogeneous equilibrium
(c) Decomposition of CaCO₃ – Homogeneous equilibrium
(d) Synthesis of HI – Homogeneous equilibrium
Ans.
(c) Decomposition of CaCO₃: Homogeneous equilibrium It is a heterogeneous equilibrium.

Question 2.

Answer:

VII. Assertion and Reason.

Question 1.
Assertion (A): Chemical equilibrium is in a state of dynamic equilibrium.
Reason (R): At equilibrium the forward and backward reactions are proceeding at the same rate and no macroscopic change is observed.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 2.
Assertion (A): In Haber’s process, NH₃ is liquefied and removed.
Reason (R): Because of the reaction keeps moving in the backward direction.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 3.
Assertion (A): In the dissociation of PCI₅ at constant pressure and temperature addition of helium at equilibrium increases the dissociation of PCl₅.
Reason (R): Helium removes CI, from the field of action.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is the not the correct explanation of A.
(c) A is true but R is false
(d) Both A and R are false
Answer:
(d) Both A and R are false

VIII. Choose the incorrect statement.

Question 1.
(a) In equilibrium mixture of ice and water kept in perfectly insulted flask, mass of ice and water does not change with time.
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate
(c) On addition of catalyst the equilibrium constant value is not affected
(d) Equilibrium constant for a reaction with negative .H value decreases as the temperature increases.
Answer:
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate

Solution:
Oxalate ions of oxalic acid form complex with ferric ions thus decrease its concentration thus, concentration of red complex in product decreases.

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium 2 Mark Questions and Answers.

I. Write brief answer to the following questions:

Question 1.
Define the state of equilibrium.
Answer:
At a particular stage, the rate of the reverse reaction is equal to that of the forward reaction indicating a state of equilibrium.

Question 2.
What are the different types of equilibrium? Explain with example?
Answer:
1. Physical equilibrium:
A system in which the amount of matter constituting different phases does not change with time is said to be in physical equilibrium.
H₂O(s) \(\rightleftharpoons\) H₂O(1). Solid-liquid equilibrium.

2. Chemical equilibrium:
Chemical reactions in which the forward and backward reactions are proceeding at the same rate and no macroscopic change is observed is said to be in chemical equilibrium. H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)

Question 3.
Explain about the equilibrium involving dissolution of solid in liquid with suitable example.
Answer:
When sugar is added to water at a particular temperature. it dissolves to form sugar solution. When more sugar is added to that solution, a particular stage sugar remains as solid and results in the formation of saturated solution. Here a dynamic equilibrium is established between the solute molecules ii the solid phase and in the solution phase.

Question 4.
How is a gas – solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecuLes in the gaseous state and those dissolved in the liquid. Example – In carbonate beverages the following equilibrium exists.
CO₂(g) \(\rightleftharpoons\)  CO₂ (Solution)

Question 5.
What is meant by active mass? Give its unit.
Answer:
The active mass represents the molar concentration of the reactants (or) products.

Question 6.
Show that K_p = K_C with two example
Answer:

Question 7.
Give two examples of equilibrium reactions where K_p > K_C.
Answer:

Question 8.
When will be K_p < K_C? Give two example.
Answer:

Question 9.
Write the K_C for the reaction CO₂(g) + H₂O(l) \(\rightleftharpoons\) H⁺_(aq) + HCO_3(aq)^–
Answer:
H₂O(l) is a pure liquid and its concentration remains constant.
Solution:

Question 10.
What is the value of K₄ in A \(\rightleftharpoons\) D
Answer:

Question 11.
Write the K_p and K_C for the following reactions
(i) 2SO₂(g) + O₂(g) \(\rightleftharpoons\) 2SO₃(g)
(ii)2CO(g) \(\rightleftharpoons\) CO₂(g) + C(s)
Answer:

Question 12.
Explain how the equilibrium constant K. predict the extent of a reaction.
Answer:
1. The value of equilibrium constant K_C tells us the extent of the reaction i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.

2. A large value of K_C indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of K_C indicates that the reaction reaches equilibrium with low product yield.

3. If K_C > 10³ the reaction proceeds nearly to completion.

4. If K_C < 10^-3 the reaction rarely proceeds.

5. If the K_C is in the range 10^-3 to 10³ significant amount of both reactants and products are present at equilibrium.

Question 13.

Predict the extent of the above two reactions.
Answer:
In the reactions, decomposition of water at 500 K and oxidation 01 nitrogen at 1000 K, the value of K_C is very less K_C < 10^-3. So reverse reaction is favoured.
∴ Products << reactants

Question 14.
Explain about the extent of reaction of dissociation of bromine mono chloride at 1000 K.
Answer:
2BrCl(g) \(\rightleftharpoons\) Br₂(g) + Cl₂(g) K_C = 5
10^-3 <  K_C <  10³
both forward and backward reaction make significant progress. Neither forward nor reverse reaction predominates.

Question 15.
What is the K_C value for formation of HCl at 700 K? Predict the extent of the reaction?
Answer:
H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g) at 700 K
K_C = 57.0
10^-3 <K_C <10³
So both forward and backward reaction make significant progress. Neither forward nor reverse reaction predominates.

Question 16.
What is the K_C value of formation of HCI at 300 K? Explain it.
Answer:
H₂(g) – Cl₂(g) \(\rightleftharpoons\) 2HCI(g) at 300 K
K₂ = 4 x 10³¹
K_C > 10³ So[products] >> [Reactant]
Reaction nearly goes to completion. So forward reaction is favoured.

Question 17.
2CO(g) + O₂(g) \(\rightleftharpoons\) 2CO₂(g) at 1000 K. What is the Kc this reaction? Predict the extent of this reaction.
Answer:
2CO(g) + O₂(g) \(\rightleftharpoons\) 2CO₂(g) at 1000 K
K_C = 2.2 x 10²²
K_C > 10³.
So [Products] >> [Reactants]
Reaction nearly goes to completion and forward reaction is favoured.

Question 18.
Define Q value for a chemical equilibrium reaction.
Answer:
Consider a homogeneous reversible reaction xA + yB \(\rightleftharpoons\) lC + mD For the above reaction under non-equilibrium conditions, reaction quotient Q is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants. Under non equilibrium conditions,
Q = \(\frac { d(Ink) }{ dt }\) ΔH° = \(\frac { { \triangle H }^{ 0 } }{ { RT }^{ 2 } }\)

Question 19.
Explain the diagrammatic expression about the direction of reaction.
Answer:

In (i) Q < K_C, the reaction will proceed in forward direction.
In (ii) Q = K_C, the reaction is in equilibrium state.
In (iii) Q> K_C, the reaction will proceed in the reverse direction.

Question 20.
Explain about the effect of catalyst in an equilibrium reaction?
Answer:
Addition of catalyst does not affect the state of equilibrium. The catalyst increases the rate of both the forward and reverse reactions to the same extent. Hence it does not change the equilibrium composition of the reaction mixture.

Question 21.
For the following equilibrium. K_C = 6.3 x 10¹⁴ at 1000K
NO(g) + O₃(g) \(\rightleftharpoons\) NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions what is K_C for the reverse reaction?
Answer:
For the reverse reaction

Question 22.
Explain why pure liquids and solids can be ignored while writing the value of equilibrium constants.
Answer:
This is because molar concentration of a pure solid or liquid is independent of the amount present.

Since density of pure liquid or solid is fixed and molar mass is also fixed, therefore molar concentration are constant.

Question 23.
A sample of Hl(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium?
2HI(g) \(\rightleftharpoons\) H₂ (g) + I₂(g)
Answer:
pHI = 0.04 atm. pH₂ = 0.08 atm; pl₂ = 0.08 atm

Question 24.
The equilibrium constant expression for a gas reaction is.

Write the balanced chemical equation corresponding to this expression.
Answer:
Balanced chemical equation for the reaction is
4 NO(g) + 6 H₂O(g) \(\rightleftharpoons\) 4NH₃(g) + 5O₂(g)

Question 25.
Predict which of the following will have appreciable concentration of reactions and products:

Answer:
Following conclusion can be drawn from the values of
(a) Since the value of K is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K is 1.8, this means that both the products and reactants have appreciable concentrations.

Question 26.
Write the equilibrium constant (K_C) expression for the following reactions.

Answer:

Question 27.
For the equilibrium 2NOCI(g) \(\rightleftharpoons\) 2NO(g) + CI₂(g) the value of the equilibrium constant K_C is 3.75 x at 1069 K. Calculate the K_p for the reaction at this temperature?
Answer:
We know that K_p = K_c(RT)^∆n_g
For the above reaction, ∆n_g = (2 + 1) – 2 = 1
K_p = 3.75 x 10^-6 (0.0831 x 1069) = 3.3 x 10^-4

Question 28.
The value of Kc for the reaction 2A \(\rightleftharpoons\) B + C is 2 x 10^-3. At a given time, the composition of reaction mixture is [A] [B] [C] = 3 x 10^-4 M. In which direction the reaction will proceed?
Answer:
For the reaction, the reaction quotient Q is given by Q_C = [B] [C]/[A]²
as [A] = [B] = [C] = 3 x 10^-4 M
Q_c (3 x 10^-4)(3 x 10^-4)/(3 x 10^-4)² = l
as Q_C > K_C so, the reaction will proceed in the reverse direction.

Question 29.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH₄(g) + H₂O(g) \(\rightleftharpoons\) CO(g) + 3H₂(g)
(a) Write an expression for K_p for the above reaction.
(b) How will the values of K_p and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst
Answer:

(b) (I) value of K_p will not change, equilibrium will shift in backward direction.
(ii) value of K_p will increase and reaction will proceed in forward direction.
(iii) no effect.

II. Answer the following questions:

Question 1.
Explain about the formation of solid-liquid equilibrium with suitable example.
Answer:
1. Consider melting of ice in a closed container at 273 K. This system reach a state of physical equilibrium in which the amount of water in the solid phase and liquid phase does not change with time.

2. In this process. the total number of water molecules leaving from and returning to the solid phase at any instant are equal.

3. If some ice cubes and water are placed in a thermos flask (at 273K and I atm) then there will be no change in the mass of ice and water.

4. At equilibrium: Rate of melting of ice Rate of freezing of water
H₂O(l) \(\rightleftharpoons\) H₂O(g)

Question 2.
How is liquid – vapour equilibrium exist?
Answer:
1. Liquid water is in equilibrium with vapour at 373 K and I atm pressure in a closed vessel

2. Rate of evaporation = Rate of condensation

Question 3.
What is meant by boiling point and condensation point of the liquid?
Answer:
The temperature at which the liquid and vapour phases are at equilibrium is called the boiling point and condensation point of the liquid.

Question 4.
Define melting point (or) freezing point of the substance.
Answer:
The temperature at which the solid and liquid phases of a substance are at equilibrium is called the melting point or freezing point of substance.

Question 5.
Illustrate the formation of solid – vapour equilibrium with suitable example.
Answer:
1. Consider a system in which the solid sublimes to vapour. e.g., I, (or) camphor.
2. When solid iodine is placed in a closed transparent vessel, after sometime, the vessel gets filled up with violet vapour due to sublimation of iodine.
3. Initially the intensity of the violet colour increases, after some time it decreases and finally it becomes constant as the following equilibrium is attained.
I₂(s) \(\rightleftharpoons\) I₂ (g)

Question 6.
Give three examples for solid vapour equilibrium.
Answer:
I₂(s) \(\rightleftharpoons\) I₂ (g)
Camphor (s) \(\rightleftharpoons\) Camphor (g)
NH₄CI(s) \(\rightleftharpoons\) NH₄CI(g)

Question 7.
Explain the following diagrams.
Diagram – 1

Answer:

1. As the concentration of the products increases, more products collide and react in the backward direction.
2. As the rate of the reverse reaction increases, the rate of the forward reaction decreases.
3. Eventually the rate of both reactions becomes equal.

Diagram-II

1. Concentration of reactants decreases with time initially and concentration of products increases with time.
2. After sometime, equilibrium is reached i.e., concentration of reactants and products remains constant.

Question 8.
What are the types of chemical equilibrium? Explain with suitable example.
Answer:

1. Chemical equilibrium is of two types:
   • Homogeneous equilibrium
   • Heterogeneous equilibrium.
2. In a homogeneous equilibrium, all the reactants and products are in the same phase
   H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
   In the above equilibrium H₂, I₂ and HI are in the gaseous state.
3. If the reactants and products of a reaction in equilibrium are in different phases, then it is calLed heterogeneous equilibrium.
   e.g., CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₃(g)

Question 9.
Write the value of K and K equation for CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₂(g)
Answer:
A pure solid always has the same concentration at a given temperature, as it does not expand to fill its container i.e., it has the same no. of moles of its volume. Therefore the concentration of pure solid is a constant. So the expression if K and K is K [CO₂], K = P_CO2

Question 10.
Consider the following equilibrium reaction and relate their equilibrium constants

1. N₂ + O₂ \(\rightleftharpoons\) 2NO, K₁
2. 2NO + O₂ \(\rightleftharpoons\) 2NO₂, K₁
3. N₂ + 2O₂ \(\rightleftharpoons\) 2NO₂, K₃

Answer:

Question 11.
Explain the effect of concentration in an equilibrium state?
Answer:
At equilibrium, the concentration of the reactants and products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentration.

According to Le – Chatelier’s principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance. For example,
H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)

The addition of H₂ or I₂ to the equilibrium mixture, disturbs the equilibrium. In order to minimize the stress, the system shifts the reaction in a direction where H₂ and I₂ are consumed i.e., formation of additional HI would balance the effect of added reactant.

Hence the equilibrium shifts to the right (forward direction). i.e., the equilibrium is re – established. Similarly, removal of HI (Product) also favours forward reaction. If HI is added to the equilibrium mixture, the concentration of HI is increased and system proceeds in the reverse direction to nullify the effect of increase in concentration of HI.

Question 12.
Consider the reaction, N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g). Explain the effect of pressure on this equilibrium reaction.
Answer:
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
In the above equilibrium, if the pressure is increased, the volume wiLl decreases. The system responds to this effect by reducing the number of gas molecules. i.e., it favours the formation of ammonia. If the pressure is reduced, the volume will increases. It favours the decomposition of ammonia.

Question 13.
Why pressure has no effect on the synthesis of HI?
Answer:
(i) When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium.
H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with ∆n_g = O.

Question 14.
Explain the effect of temperature on the following equilibrium reaction.
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) ∆H = – 92.2 kJ.
Answer:
In this equilibrium, N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) ∆H = – 92.2 kJ.

Forward reaction is exothermic while the reverse reaction is endothermic. If the temperature of the system increased, the system responds by decomposing some of ammonia molecules to nitrogen and hydrogen by absorbing the supplied heat energy.

Similarly, the system responds to a drop in the temperature by forming more ammonia molecule from nitrogen and hydrogen which release heat energy.

Question 15.
How does oxygen exchanges between maternal and fetal blood in a pregnant woman?
Answer:
1. In a pregnant women, the oxygen supply for the fetus is provided by the maternal blood in the placenta where the blood vessels of both mother and fetus arc in close proximity. Both fetal and maternal hemoglobin binds tO oxygen reversibly as follows.

2. In the above two equilibrium, the equilibrium constant value for the oxygenation of fetal hemoglobin is higher which is due to its higher affinity for oxygen compared to adult hemoglobin. Hence in placenta, the oxygen from the mother’s blood s effectively transferred to fetal hemoglobin.

Question 16.
What is K for the following reaction in state of equilibrium?
2SO₂(g) + O₂ \(\rightleftharpoons\) 2SO₃(g)
(Given:[SO₂] = 0.6 M; [O₂] = 0.82 M; and [SO₃] = 1.90 M
Answer:
2SO₂(g) + O₂ \(\rightleftharpoons\) 2SO₃(g)

Question 17.
At a certain temperature and total pressure of 10⁵Pa, iodine vapours contain 40% by volume of iodine atoms in the equilibrium 1₂(g) \(\rightleftharpoons\) 2I(g). Calculate K_p for the equilibrium.
Answer:
According to available data:
Total pressure of equilibrium mixture = 10⁵ Pa
Partial pressure of iodine atoms (1) = \(\frac { 40 }{ 100 }\) x (10⁵ Pa) = 0.4 x 10⁵ Pa
Partial pressure of iodine molecules (I₂) = \(\frac { 60 }{ 100 }\) x (10⁵ Pa) = 0.6 X 10⁵ Pa

Question 18.
A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant K_C for the reaction
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) is 1.7 x 10^-2
Is this reaction at equilibrium? if not. what is the direction of net rection?
Answer:
The reaction is: N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
Concentration quotient (Q ) =

The equilibrium constant (Kr) for the reaction = 1.7 x 10^-2
As Q_C \(\neq\) K_C; this means that the reaction is not in a state of equilibrium.

Question 19.
What is the effect of:

1. addition of H₂
2. addition of CH₃OH
3. removal of CO
4. removal of CH₃OH

On the equilibrium 2H₂(g) + CO(g) \(\rightleftharpoons\) CH₃OH(g)
Answer:

1. Equilibrium will be shifted in the forward direction.
2. Equilibrium will be shifted in the backward direction.
3. Equilibrium will be shifted in the backward direction.
4. Equilibrium will be shifted in the forward direction.

Question 20.
At 473 K, the equilibrium constant K_c for the decomposition of phosphorus pentachioride (PCl₅) is 8.3 x 10^-3 . If decomposition proceeds as:
Answer:
PCI₅(g) \(\rightleftharpoons\) PCI₃(g) + Cl₂(g);
∆H = + 124.0 kJ mol^-1

1. Write an expression for K. for the reaction.
2. What is the value of K for the reverse reaction at the same temperature.
3. What would be the effect on K_C if
   • More of PCI₅ is added
   • Temperature is increased.

Answer:
1. The expression for
2.  For reverse reaction
3.
(I) By adding more of PCI₅, value of K^c will remain constant because there is no change in temperature.
(ii) By increasing the temperature the forward reaction will be favoured since it is endotherniic in nature. Therefore, the value of equilibrium constant will increase.

Question 21.
Dihydrogen gas used in Haher’s process is produced by reacting methane from natural gas with high temperature stam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
Answer:
CO(g) + H₂O(g) \(\rightleftharpoons\) CO₂(g) + H₂(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that P_CO = P_H2O = 4.0 bar., what will be the partial pressure of 2 at equilibrium? K_p = 0.1 at 400°C.
Answer:
Let the partial pressure of hydrogen (H₂) at equilibrium point p bar

Question 22.
The value of K_C for the reaction 3O₂(g) \(\rightleftharpoons\) 2O₃(g) is 2.0 x 10^-50 So at 25°C. If equilibrium concentration of 0, in 25°C is 1.6 x 10² what is the concentration of O₃?
Answer:
3O₂(g) \(\rightleftharpoons\) 2O₃(g)

Question 23.
The reaction CO(g) + 3H₂(g) \(\rightleftharpoons\) CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1K flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_C for the reaction at the given temperature is 3.90.
Answer:

Question 24.
The following concentration were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium at 500 K.

• [N₂(g)] 1.5 x 10^-2 M
• [H₂(g)] = 3.0 x 10^-2 M
• [NH₃]= 1.2 x 10^-2M.

Calculate equilibrium constant.
Answer:
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
Calculate equilibrium constant
Answer:

Question 25.

1. In the reaction A+ B → C + D, what will happen to the equilibrium if concentration of A is increased?
2. The equilibrium constant for a reaction is 2 x 10^-23 at 25°C and 2 x 10^-2 at 50°C. Is the reaction endothermic or exothermic?
3. Mention at least three ways by which the concentration of SO₃ can be increased in the following reaction in a state of equilibrium. 2SO₂(g) + O₂(g) \(\rightleftharpoons\) 2SO₃(g)

Answer:

1. The reaction will shift in the forward direction.
2. Endothermic
3. following reaction
   • increasing concentration of SO₂
   • increasing pressure.
   • increasing concentration of oxygen.

Question 26.
PCI₅, PCI₃ and CI₂ are at equilibrium at 500 K and having concentration I.59M PCl₃, l.59M
CI₂ and 1.41M PCI₅. Calculate K. for the reaction PCl₅ \(\rightleftharpoons\) PCl₃ + Cl₂
Answer:
The equilibrium constant K. for the above reaction can be written as:

Question 27.
Given the equilibrium
N₂O₄(g) \(\rightleftharpoons\) 2NO₂(g) with K_p = 0.15 atm at 298 K
(a) What is K_p using pressure in torr?
(b) What is K_C using units of moles per litre.
Answer:

III. Answer the following questions in detail:

1. Derive the values of equilibrium constants K_p and K_C for a general reaction
x A + y B \(\rightleftharpoons\) lC + mD
Answer:
Let us consider a reversible reaction x A + y B \(\rightleftharpoons\) lC + mD
where A, B are the reactants C and D are the product and x, y. l and m are the stoichiometric coefficients of A. B, C and D respectively. Applying the law of mass action the rate of forward reaction.
r_f ∝ [A]^x [B]^y or r_f K_f [A]^x [B]^y
Similarly the rate of backward reaction
r_b ∝ [C]^l [D]^m or r_b = K_b [C]^l [D]^m
where K_f and K_b are proportionality constants.
At equilibrium, Rate of forward reaction (r_f) = Rate of backward reaction (r_b)

where K_C is the equilibrium constant in terms of concentration. At a given temperature, the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants is a constant known as equilibrium constant.

If the reactants and products of the above reaction are in gas phase, then the equilibrium constant can be written in terms of partial pressures

where P_A, P_B, P_C and P_D are the partial pressure of gases A, B, C and D respectively.

Question 2.
Derive the values of K and K for the synthesis of HI.
Answer:
H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
Let us consider the formation of HI in which ‘a’ moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’. Let ‘x’ moles of each of H₂, and I₂, react together to form 2x moles of HI.

Question 3.
Derive the values of K and K for dissociation of PCI₅.
Answer:
Consider that ‘a’ moles of PCl₅ is taken in container of volume ‘V’ Let x moles of PCI₅ be dissociated into x moles of PCI₃ and x moles of Cl₂.

Applying law of mass action

Where ‘n’ is the total number of moles at equilibrium

Question 4.
At certain temperature and under a pressure of 4 atm, PCl₅ is 10% dissociated. Calculate the pressure at which PCI₅ will be 20% dissociated at temperature remaining constant.
Calculation of K_p

Total no. of moles in the equilibrium mixture = 1 – α + α + α = (1 + α) mol.
Let the total pressure of equilibrium mixture = ρ atm



calculation of P under new condition

IV. Numerical Problems

Question 1.
Find the value of K for each of the following equlibria from the value of K
(a) 2NOCI(g) \(\rightleftharpoons\) 2NO(g) + Cl₂(g); K_p= 1.8 x 10² atm at 500 K
(b) CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₂(g): K_p = 167 atm at 1073 K.
Answer:

Question 2.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICI was 0.78 M?
2ICI(g) \(\rightleftharpoons\) I₂(g) + CI₂(g); K_C = 0.14
Answer:
Suppose at equilibrium, the molar concentration of both I₂(g) and Cl₂(g) is x mol L^-1.

Question 3.
Equilibrium constant K for the reaction. N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) at 500 K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is 3.0 mol L^-1 of N₂; 2.0 mol L^-1 of H₂; 0.50 moI L^-1 of NH₃. Is the reaction at equilibrium?
Answer:
The given reaction is: N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
According to available data

Common Errors

1. In writing K_p, K_C values from the equations, students may confuse to write whether products or reactants in the numerator or in the denominator
2. When K_p = K_C.(RT)⁰ students may confuse.
3. In writing An values, students will consider all reactants and products.
4. Students are confused to understand the concept of Q value and K_C value.
5. In writing chemical equilibrium reaction, you may miss the physical states of reactants and products.
6. ∆n_gvalue calculation will go wrong if you consider all the reactants and products.
7. Equilibrium constant value is calculated under equilibrium condition and reaction quotient value defined at sometimes at equilibrium condition wrongly by the students.
8. Chemical equilibrium condition must be known. Students may wrongly write in an open vessel, the equilibrium take place.
9. Equilibrium symbol, students may wrongly write as =

Rectifications

1. Always we have to write K_p & K_C as
2. (Anything)° = 1. So K_p = K_C.
3. Only they have to consider gaseous reactants and gaseous products. Since solid and liquid are constant in their concentration.
4. Q value is calculated under non equilibrium conditions. K_C value is calculated under equilibrium condition.
5. Physical states of reactants and products must be written as a subscript by the words s, l, g. (solid, liquid, gas)
6. ∆n_g = n_gp – n_gr You should consider only gaseous products and gaseous reactants.
7. Reaction quotient Q value is calculated only under non-equilibrium conditions.
8. Chemical equilibrium reactions are always take place at closed vessel only.
9. Equilibrium reaction symbol is ⇔.

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Tamilnadu Samacheer Kalvi 11th English Letter Writing

Check out the topics covered in Writing Letter Writing Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Writing Letter Writing Questons and Answers. This helps to improve your communication skills.

Letters are the most common forms of written communication. Writing letters is an art and it is mastered through practice. We write letters to friends and relatives to maintain contacts with them. However, in the present times with the boom in methods of communication, many of us don’t have the time or the inclination, or the temperament and the art to write letters. So telephone, fax and e-mail have replaced personal (informal) letters.

 

However, formal letters are still in vogue. Commerce, trade, official correspondence, public representation, complaints and other dealings, transactions and communication with the people are still conducted through letters. Students are therefore advised to cultivate the art and skill of letter writing. It must be remembered that different kinds of letters follow different conventions. . So, the format should suit the type.

Informal Letters
These are ordinary personal letters and informal invitations written to relatives, friends and acquaintances. The main characteristics of informal letters are:

• These letters are first/second person presentations.
• Personal feelings and emotions find an. expression.
• The writer’s address is given in the usual place.
• The salutation is usually ‘Dear’ plus ‘Name’.
• The date of writing is given, but the year is generally omitted.
• The style and tone are relaxed and informal.
• Different tenses are used as the sense demands.
• The complimentary close is: ‘Yours lovingly’ or ‘Yours truly’.

 

Formal Letters

A formal letter is different from a personal letter in tone and content. These letters are written for official purposes or commercial correspondence, so they must be systematic, specific in content and formalin the treatment of the’subject. The main features of formal letters are:

• The matter is brief, clear and to the point.
• There are separate paragraphs for separate topics.
• The first paragraph indicates the theme of the letter.
• The facts are stated in simple and direct language.
• Long and high sounding words are avoided.
• Even while lodging a complaint or making criticism, the tone should be polite.
• Clumsy expressions should not be used.
• The complimentary close is: ‘Yours sincerely’, or ‘Yours faithfully’.

 

Format/Layout Of A Formal Letter

08th March, 20XX

From
XXX
Sender’s Address & Phone No.

To
THE DIRECTOR
Addressee’s Address
………………………………
………………………………

Respected Sir/Madam,
SUB: For the post of a Senior Accountant
REF: The Hindu dated 06/03/20XX

With reference to your advertisement dated 6th March, 20XX in The Hindu, I hereby wish to apply for the post of a Senior Accountant in your esteemed organization.

If given an opportunity I assure you of my sincere service to the satisfaction of my superiors.

Please find enclosed my Bio-data for your kind perusal and looking forward to a positive reply at the earliest.

 

Thank you,
Yours sincerely,
XXX

To
The Director
………………………………
………………………………

Note: There is no punctuation after From and To
Thank you has been used instead of Thanking you

Types Of Formal Letters

A. Business Letters

• Making enquiries/asking for information
• Replying to enquiries/giving information
• Placing orders and sending replies
• Cancelling orders
• Letters of complaints

B. Official Letters

• Registering complaints regarding civic amenities, law & order, etc.
• Making enquiries pertaining to deveopmental projects, health facilities, etc.
• Making requests/appeals

 

C. Letter To The Editor

• Giving suggestions on an issue (usually of public interest)
• Expressing views on an issue already raised in an article/write-up/in a published letter

D. Letter Of Application

• Applications for jobs

What Is A Bio-Data?
The application for a job is sometimes accompanied by a bio-data of the applicant. The bio-data is also called a resume or curriculum vitae (C.V.). It contains the following information about the candidate:

• Personal information, e.g., age, health, sex, etc.
• Educational/professional qualifications
• Experience/suitability for the job

 

Bio-Data
Name: Xxx
Date Of Birth: 8th May, 1986
Marital Status: Married.
Husband’S Name: Mr. Shandy Rajeevan
Address For ComMUNICATION: YYY
Contact Number – MOBILE: 9988776655
Residence: 01222445566
Mother Tongue: Tamil
Language Known: English and Tamil, Hindi And Malayalam (to speak)
Passport Number: A7265511

Educational BacKGROUND : (In reverse chronological order)

Professional ExPERIENCE : (In reverse chronological order)

 

Hobbies: Photography, Gardening, Reading, Travelling
Expected Salary: Rs 35,000/per month
Salary Drawn: Rs 32,000/ per month
Reference: (1) Mr. Ravi (XYZ Pvt. Ltd.) 9998887777, (2) Mrs. Rani (Raj Enterprises) 9900000222

Declaration
I hereby declare that the above given information is true to my knowledge.
Station: YYY
Date: 08.03.20XX

XXX
SIGNATURE OF THE APPLICANT

Solved Questions
(i) Write a letter to your father seeking his permission to join the swimming class.
Answer:
Chennai
01.05.17

Dear father,

I’m fine. I hope all are fine at home. I wish to join a swimming class. Mr. Rakesh is the coach. The charges are only two hundred rupees per month. The classes start next Monday. Please give me permission.

Thank you,

Yours lovingly,
XYZ

 

To
Mr. Somu,
12,” Railway Station Road,
Thirunelveli

(ii) As Sports Secretary of your school, write a letter to the Secretary of the Sports Authority of India, Delhi requesting him for details regarding the sports scholarships that are available for students.
Answer:
20 April, 20XX
S. Narula
St. John’s School
Ramapuram
Chennai

The Secretary
Sports Authority of India
New Delhi

Sir,
Sub: Information regarding sports scholarship
I have come to know from my friend in Hyderabad that the Sports Authority of India awards scholarships to sports persons and athletes who distinguish themselves in various sports disciplines during their school days. The tenure of these sports scholarships varies from three to five years.

 

Some of our athletes have topped at the district level and have been selected for national events. They are keen to know the availability of sports scholarships awarded under your benign guidance. The specific conditions and eligibility criteria may please be intimated to us.

I hope you will provide the necessary information at the earliest.

Yours faithfully

S. Narula
Sports Secretary

To
The Secretary
Sports Authority of India
New Delhi

 

(iii) Ravi Mohan of 59 Adarsh Nagar, Secunderabad, a Class XII student of the Commerce stream sees this advertisement.
Coaching in Commerce Stream
Sure Shot Institute
14-A Lancer Barracks, Secunderabad
The best choice to ensure success in the exams
Hurry! Join our classes! Limited seats

He writes a letter to the institute seeking information about the subjects taught, the timings of the classes, class size and fees. Write his letter.
Answer:
16 July, 20XX
Ravi Mohan
59, Adarsh Nagar
Secunderabad
The Director
Sure Shot Institute
14 A, Lancer Barracks
Secunderabad Sir,

 

Sub: Coaching in Commerce stream
Kindly refer to your advertisement in the Deccan Times dated 15 July. I would like more information and details about the courses the institute offers, facilities available, timings, fee structure and size of classes/groups. I need coaching in Accountancy, Advanced Mathematics, and Statistics. Please let me know if coaching is available in all these subjects and what is the schedule. Does your institute cater to individual difficulties or do you discuss and resolve general problems?

Kindly enlighten me on all the above points. In case you have a detailed information booklet, do send it to my address given above.

Yours faithfully
Ravi Mohan

To
The Director
Sure Shot Institute
14 A, Lancer Barracks
Secunderabad

 

(iv) Write a letter to the Police Commissioner (Traffic) about the inadequate parking facilities in the commercial area of Nungambakkam, which is causing a lot of inconvenience to the people. You may also offer your suggestions to solve it. You are Rakesh/Radhika, No. 12, Nungambakkam, Chennai. (Word limit: 150 words) .
Answer:
12 March, 20XX
Radhika
No. 12, Nungambakkam
Chennai

The Police Commissioner (Traffic)
Chennai
Sir,

Sub: Inadequate parking facilities

 

I wish to draw your attention towards the lack of an essential civic amenity which causes trouble not only to the vehicle users but also to the general public. The inadequate parking facilities in the commercial street area of Nungambakkam has caused a sea of chaos, confusion and disorder. Most of the vehicles are parked on roads blocking the passages to and from the parking spaces. The problem gets worse during the evening hours. People are stranded as they can’t park their vehicles at places earmarked for parking. Nor can they move out easily towards the road from the parking spaces.
I would like to offer some practical suggestions:

(i) Vehicles with even number should be allowed on even days, i.e., (Tuesday, Thursday F and Saturday) while those with odd ones be used on odd days (Monday, Wednesday, Friday). This will cut the number of vehicles by 50%.

(ii) The parking lots should be controlled by the police so that entry/exit is not blocked.
I hope my suggestions will be implemented for public benefit.

Thank you,
Yours faithfully
Radhika

To
The Police Commissioner (Traffic)
Chennai

 

(v) As the Regional Manager of a leading XYZ Textile Company, Madurai, write a letter to the American Textile Company, Chennai offering two of your products for bulk sale.
15 April, 20XX
Xyz Textile Company
24 Temple Road
Madurai

The Senior Purchase Manager ,
The American Textile Company
14 Tank Bund Road
Chennai
Sir,
Sub: Offer of Products for Bulk Sale
Ref. STC/20/78

We are one of the leading manufacturers of textile products in India and our products are equally popular at home and abroad. We have received an enquiry from your branch office at Chennai regarding terms for bulk purchase.

We wish to inform you that we grant 30% trade discount to commercial/bulk purchases on an order for not less than one lakh at a time. We allow further discount of 5% to buyers whose orders exceed Five lakh per year.

Our latest price list and illustrated catalogue along with specimen of textiles are enclosed for your kind perusal and approval.
An early reply will be appreciated.

 

Yours sincerely
A. G. George Sales
Manager

To
The Senior Purchase Manager
The American Textile Company
14 Tank Bund Road
Chennai

(iv) You have placed an order for a few books with City Central Book Shop, Chennai. You have not received the books so far. Write a letter to the bookshop complaining about the non-compliance of your order.
Answer:
25 March, 20XX
A. Shanmugan
Globe Sr. Sec. School
Adambakkam
Chennai

City Central Book Shop
3rd Street
Mount Road
Chennai

 

Sir,
Sub: Non-compliance of the order No. 11/RS

This is with reference to our order No. 11/RS dated 4th March, 20XX regarding the delivery of a few books for our school library.

In this respect, I regret to state that despite repeated reminders on the phone, the books have not yet been delivered to the library so far. The new session has started and the books are required urgently for reference work by the students. I am quite upset about the non-compliance and negligent attitude shown by your shop towards our order. If the specified books don’t reach us within three days from the receipt of this letter, we will conclude that you are not interested in fulfilling this order. In that case, we’ll be compelled to place the order elsewhere. A list of

books ordered is being enclosed herewith. Kindly do the needful at the earliest.

Yours truly
A. Shanmugan
(Librarian)
Enel.—Photocopy of the book-list.

To
City Central Book Shop
3rd Street
Mount Road
Chennai

 

(vii) You are Vasanthi, a resident of Ayanavaram, Chennai. The residents feel inconvenience due to the frequent digging up of roads by various departments such as electricity, water supply, sewage, telephone, etc. Write a letter to the Editor, The Times of India, Chennai, highlighting the problem and suggesting remedial measures.
Answer:
25 July 20XX
Vasanthi
No. 2/8, Ayanavaram

The Editor
The Times of India
Chennai

Sir,
Sub: Frequent Digging up of Roads

I would like to highlight the problems faced by the residents of Ayanavaram due to the frequent digging up of roads by various departments, such as electricity, water supply, sewage, telephones, etc. We have to put up with traffic diversions almost every fortnight. The dug roads remain unrepaired and cause traffic hazards as well as accidents.

We have written to the Development Authority many times, but our repeated efforts have failed to bear any fruit. I hope the publication of the letter in your newspaper will draw the attention of the authorities to our plight and motivate them to expedite the repair work.

 

Yours faithfully
Vasanthi

To
The Editor
The Times of India,
Chennai

(viii) Ready Assurance Company, Coimbatore has given an advertisement in ‘The Hindu’ for recruitment of management trainees to be groomed as managers of their company. Apply for the same, giving your detailed bio-data (curriculum vitae). You are Asokan/ Adithi, 5th Main Road, T. Nagar, Chennai.
Answer:
10 March 20XX
Asokan
5th Main Road, T. Nagar
Chennai

The Personnel Manager
Ready Assurance Company
Coimbatore

Sir,
Sub: Recruitment of Management Trainee

 

With reference to your advertisement in The Hindu dated 5th March 20XX for management trainees to be groomed as managers for your company, I would like to be considered for the said post.

My Bio-data is enclosed for your perusal and consideration. If I am found suitable, I can appear for the interview at any time suitable to you. In case of selection, I assure you of my unstinted cooperation and devotion in the discharge of my duties.

Yours faithfully,
Asokan

To
The Personnel Manager
Ready Assurance Company
Coimbatore

Bio-Data
Name : Asokan Rajan
Father’s Name : Prof. S. Rajan
Address : 5th Main Road, T. Nagar, Chennai
Date of Birth : 7th August 1987
Educational Qualifications : 1. B.Com., Madras University, 2. MBA from Madurai
Experience : 2 years
Present Employment : Working with Global Management, Chennai
Marital Status : Unmarried
Personal Details : Age 29 years, Height : 5.5, Weight : 60 kg
Languages known : Tamil, English, Hindi, French
Hobbies : Reading, Listening to music, Painting
Reference : Professor Suresh Mohan, Madras University

 

Declaration
I hereby declare that the above given information is true to my knowledge.
Station : YYY
Date : 10.03.20XX

Asokan

(ix) You are Sunil/Sujatha Kannan, parent of a student of class XI. Write a letter to the Principal, Brilliant Public School, Dharmapuri, drawing his attention to the disorderliness, non-availability of books and lack of cooperation from the library staff and suggesting some improvements in the functioning of the school library.
Answer:
25 March 20XX
Sujatha Kannan
37, Dharmapuri
Tamil Nadu

The Principal
Brilliant Public School
Dharmapuri
Sir,

 

Sub: Improvements in School Library

During my visit to your school in connection with P.T.A. meeting, I happened to go to the library. I would like to draw your attention to the malfunctioning of the library. I sincerely feel that it can prove more useful to the students if few steps are taken to tone it up.

I was shocked at the disorderliness in the library. Books were lying in heaps and were not restored to the proper shelves. No wonder the book which the library has is reported to be non-available by the library staff. The lack of co-operation from the library staff makes the situation worse. You may take steps to make them alert and willing workers. The shortage/ non-availability of books, if any, may be made up by buying more copies of the volumes more in demand. The old and broken tables and chairs may be repaired if not replaced.

I hope that during our next meeting we shall find the library cosy, spick and span.

Yours faithfully
Sujatha Kannan

To
The Principal
Brilliant Public School
Dharmapuri

(x) Write a letter to the Headmaster of your school requesting him to issue your Transfer Certificate.
Answer:
26 July 20XX
S. Aran
12, Nehru Street
Alanganallur
The Headmaster
Government Boy’s Hr. Sec. School
Alanganallur

 

Sir,
Sub: Request to issue Transfer Certificate
I completed my Std X examination recently. My Roll No. was 18.1 studied in ‘B’ section. I am going to apply for ITI course pending the publication for SSLC results. Kindly issue me my Transfer Certificate.

Thank you,
Yours sincerely
S. Arun

To
The Headmaster
Government Boy’s Hr. Sec. School
Alanganallur

 

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Writing Letter Writing Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

Students who are interested in learning of 11th English Writing Narrative Writing Questions and Answers can use Tamilnadu State Board Solutions of 11th English Chapter Wise Pdf. First check in which chapter you are lagging and then Download Samacheer Kalvi 11th English Book Solutions Questions and Answers Summary, Activity, Notes Chapter Wise. Students can build self confidence by solving the solutions with the help of Tamilnadu State Board English Solutions. English is the scoring subject if you improve your grammar skills. Because most of the students will lose marks by writing grammar mistakes. So, we suggest you to Download Tamilnadu State Board 11th English Solutions according to the chapters.

Tamilnadu Samacheer Kalvi 11th English Narrative Writing

Check out the topics covered in Writing Narrative Writing Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Writing Narrative Writing Questons and Answers. This helps to improve your communication skills.

Under this head you may be asked to recount an event/incident that you have experienced. It is a first-person account of the event or incident.

 

Format: Heading Writer’s name and class
Language: Should be semi-formal. Try to be simple but attractive and appealing. Avoid displaying your linguistic ability.

Content :

– factual information about incident/experience
– date, time, venue of the incident/experience
– sequence of actions/incidents
– reaction to the incident

Solved Questions
(i) Your grandmother completed eighty years of her age on August 16, 2017. Celebrating her 80th birthday was an event for the family.

Describe the event in your words (150-200 words) mentioning the following points:

 

– preparations for the occasion
– people who gathered
– honour given to the grandmother
– her reactions to the occasion
– her personality
– smart, witty, etc.
– your reactions

Answer:

Grandma Turns Eighty
[Naren, XI A]

Sixteenth August was a grand occasion for our family. My grandmother had turned eighty that day. We organised a family get-together. Messages had been sent to all my uncles, aunts and cousins. The ancestral home was decorated with flowers. A puja was performed in the temple. Then the main function began in the sitting room. It was a very cheerful occasion. All my uncles, aunts and cousins gathered under one roof. She was seated in a high arm chair. My uncle honoured her with a beautiful shawl. Then my parents presented her an almond coloured silk saree. Then came the turn of youngsters.

 

She appreciated all the gifts presented to her and blessed us. She is still smart, witty and energetic. Words of wit and wisdom dropped like honey from her lips. Dressed in her usual orange coloured dress, she appeared like a divine personality. Since I was the youngest member of the family, I received love and . affection from everyone. Sometimes I felt it was my birthday.

(ii) During the summer vacation Magesh visited his grandfather living in his native village Perambalur – Peraiyur. It was a very pleasant yet unusual experience for him. Thinking you are Magesh, write about such experiences in 150-200 words.
Answer:

Native Village Revisited
[by Magesh]

I had been to my native village five years ago and had very faint memories. But when I visited my grandfather during this summer vacation, all the earlier experiences were revived. What a contrast the village presented. Instead of bullocks, I saw tractors ploughing the fields. Electric motors were drawing water to irrigate fields which were full of tall and green maize plants. Many trees were planted on the boundaries of the farm.

However, the scene at the village pond remained more or less unchanged. I could still notice buffaloes lying in water, a boy riding one of them and rural women washing clothes there. What shocked me was that filthy water from drains was being carried to the pond. The old habits of villagers still persisted. Heaps of rubbish and dung were lying here and there. Stray dogs and pigs were wallowing in the dirt and mud. I beat a hasty retreat to my farmhouse as I could not stand the filth and nauseating smell.

 

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Writing Narrative Writing Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

Students who are interested in learning of 11th English Writing Report Writing Questions and Answers can use Tamilnadu State Board Solutions of 11th English Chapter Wise Pdf. First check in which chapter you are lagging and then Download Samacheer Kalvi 11th English Book Solutions Questions and Answers Summary, Activity, Notes Chapter Wise. Students can build self confidence by solving the solutions with the help of Tamilnadu State Board English Solutions. English is the scoring subject if you improve your grammar skills. Because most of the students will lose marks by writing grammar mistakes. So, we suggest you to Download Tamilnadu State Board 11th English Solutions according to the chapters.

Tamilnadu Samacheer Kalvi 11th English Report Writing

Check out the topics covered in Writing Report Writing Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Writing Report Writing Questons and Answers. This helps to improve your communication skills.

Report writing is an attempt to gather information about an event, incident or accident from the persons concerned, the parties involved, the victims and authorities. The third-person point of view ensures objectivity in the report.

 

Guidelines
The heading is essential. The report may be in one or two paragraphs.

• Be objective.
• Organize the details properly. and logically.
• Present the material systematically

How should a report be written?
Answer:
A report should:

– be in the form of a narrative
– include all relevant details
– mention the date and time of occurrence
– mention the facts
– be in the past tense
– focus on one particular event only
– mention the venue
– mention the cause, result, etc.

 

Report For A School Magazine
Format:
(a) Heading
(b) Writer’S Name And Class.

Heading/Title Of The Happening [by Dhanwanth/Aditi] ………………………………………………………………………… ………………………………………………………………………… ………………………………………………………………………… …………………………………………………………………………

Language: Should be semi-formal. Try to be simple but attractive and appealing. Avoid displaying your linguistic ability.

 

Content :

– factual information about school activity/event
– date, time, venue of the activity/event
– a sequence of event/program
– information about participants/chief guests/judges
– kind of organisation, people responsible for programme/arrangements
– results, if describing a contest

Report For A Newspaper
Usual subjects: Political news, sports news, crimes, accidents, natural disasters, etc.

Format :
(a) Headline
(b) ‘Byline’, i. e., by a correspondent/reporter or an agency as its source and
(c) Date-line-date and place of occurrence

 

Headline [by Sudhir/Sudha, TOI Correspondent/Staff reporter] Chennai, 9 March ………………………………………………………………………… ………………………………………………………………………… ………………………………………………………………………… …………………………………………………………………………

Language and Style :

– quite formal
– passive voice is preferred
– journalistic jargon (vocabulary/expressions), e.g., according to ministry/party spokesman or according to government sources
– use of words like ‘alleged’ or ‘suspected’ before ‘murderer’, ‘smuggler’, ‘thief’, etc.

 

Content:
It is most important. The main information is given in the first two or
three sentences. Other essential information like date, place, occasion, etc., follow. The presentation should be symmetrical.

Solved Questions
(a) Write a report for your school magazine.
Answer:

The Annual Sports Day
[by Shekhar]

12th February, 20XX, was a big occasion for our school. The Annual Sports of our school were held on that day at our school playground. A colorful shamiana was erected for guests and teachers. The stands were also tastefully decorated. The athletes gathered in front of the stage at 8.30 a.m. There was a march-past. The Principal took the oath and declared the sports- meet open. At 9.30 the track events began with 200 metre race for boys and 100 metre race for girls. Field events like long jump, high jump, javelin throw and discus throw were held in between the races. The programme was beautifully planned. In the afternoon the cycle race, sack race and three legged-race provided amusement. The musical chair race for guests provided a lot of fun. Kamal of XIIA was declared the best athlete.

 

(b) The floods in the State of Tamil Nadu, adversely affected the area causing destruction to life, property, cattle and crops. As a news correspondent, you visited this state. Write the details of this disaster in 150-200 words. Also give a suitable title to your write up.
Answer:

Devastation By Floods
[XYZ, News Correspondent, The Express]

A visit to the flood-hit districts of the State of Tamil Nadu fills one with depressing thoughts. How can nature be so cruel? The cyclonic storm has proved a disaster for the state. Coastal areas have been hit hard. The villages and hamlets have been adversely affected. Thousands of acres of land having crops are still submerged in water. Millions of cattle have been washed away or perished in the flood waters. Thousands of the villagers have been rendered homeless. Hundreds of children and the aged have been swept off. Thus there has been an all round destruction of life, property, cattle and crops. Government agencies and voluntary groups have rushed to the help of the flood-affected victims. The situation is grim and calls for more concerted efforts.

 

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Writing Report Writing Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

Students who are interested in learning of 11th English Prose Chapter 6 The Accidental Tourist Questions and Answers can use Tamilnadu State Board Solutions of 11th English Chapter Wise Pdf. First check in which chapter you are lagging and then Download Samacheer Kalvi 11th English Book Solutions Questions and Answers Summary, Activity, Notes Chapter Wise. Students can build self confidence by solving the solutions with the help of Tamilnadu State Board English Solutions. English is the scoring subject if you improve your grammar skills. Because most of the students will lose marks by writing grammar mistakes. So, we suggest you to Download Tamilnadu State Board 11th English Solutions according to the chapters.

Tamilnadu Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Check out the topics covered in Prose Chapter 6 The Accidental Tourist Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Prose Chapter 6 The Accidental Tourist Questons and Answers. This helps to improve your communication skills.

Warm up

Question 1.
Often on formal occasions, we admire friends and strangers who appear elegant, who are pleasant to converse with and who conduct themselves gracefully. At times, we also see people who are awkward, nervous and doubtful about their next move.
Recall a few examples of awkward actions that can cause discomfort or disturbance to others like spilling a cup of hot drink on someone nearby.
Answer:
Mr. X has travelling sickness. Once he was in a city bus, his wife advised him to take a polythene bag so that he can vomit inside it if he felt like it. But he forgot her advice. As the bus moved on, he had a feeling that he wasn’t fine. He craned his neck out of the window of the bus and vomited his undigested breakfast.

As the bus was moving fast, people seated behind him had their shirts and sarees sprinkled with the undigested breakfast and dinner. The conductor and driver got upset. The whole day, passengers refused to sit on those three seats.

Samacheer Kalvi 11th English The Accidental Tourist Textual Questions

A. Based on your understanding of the lesson, answer the following questions in one or two sentences each:

Question 1.
Give a few instances of Bryson’s confused acts.
Answer:
He would end up standing in an alley on the wrong side of a self-locking door trying to locate a restroom in a cinema theatre. He would often go to the hotel desk, atleast two to three times a day asking what was his room number. The author had a serious problem of orientation in any new place:

Question 2.
What were the contents of the bag?
Answer:
The contents of the bag were frequent flyercard, newspaper cuttings, loose papers, tobacco pipe, magazines, passport, English money and film.

 

Question 3.
Describe the Buttery cascade of things tumbling from the bag.
Answer:
The side of the bag flew open and everything within was extravagantly ejected over an area about the size of a tennis court. The carefully stored documents came raining down in a fluttery cascade. Coins bounced to a variety of noisy oblivions. Lidless tobacco tin rolled crazily across the concourse disgorging its content as it went.

Question 4.
Why did the author’s concern over tobacco shift to his finger?
Answer:
When the author saw the racing tobacco box disgorging its content, he worried about the need to buy expensive tobacco in England. But when he saw his own bleeding finger that he had gashed while forcefully opening the jammed zip, he shifted his cry to his finger.

Question 5.
What happened to Bryson when he leaned to tie his shoelace?
Answer:
When Bryson leaned to tie a shoe lace inside the air-craft, some one in the seat ahead of him threw his seat back into full recline. The author found himself pinned helplessly in a crash position.

Question 6.
How did Bryson free himself from the crash position?
Answer:
Bryson was able to disentangle himself only by clawing the leg of the man sitting next to him.

Question 7.
Give a brief account of the embarrassing situation of Bryson when he knocked down the drink.
Answer:
Bryson rose from the dinner table looking as if he had just experienced a localized seismic event. While opening the lid, he used to spill the contents on his family and fellow passengers and thus embarrassing them. To avoid such anxious moments, his wife used to give directions to the children. “Take the lids off the food for daddy”.

Question 8.
What was Bryson’s worst accident on a plane?
Answer:
During one of the flights, the author was jotting down his thoughts on a notebook. The important thoughts reflected how well he behaves in public places. His recorded thoughts were, “buy socks”, “clutch drinks carefully”. He was sucking the pen thoughtfully. He was oblivious of the fact that the pen was leaking. The leaked ink had left scrub-resistant navy blue on his mouth, gum, chin, tongue and teeth for several days.

Question 9.
What did Bryson wish to avoid in his life?
Answer:
Bryson wished to avoid air travel especially with his family members.

Question 10.
How would staying away from liquid mischief benefit Bryson?
Answer:
Staying away from liquid mischief would naturally reduce the expense on laundry bills.

B. Answer the following questions in about three to four sentences each:

Question 1.
Why doesn’t Bryson seem to be able to do easily what others seem to? Give a few reasons.
Answer:
Bryson lacks orientation. He has. great capacity in forgetting hotel rooms, the location of rest rooms in a cinema hall and the number of his room in a hotel. He goes atleast three times a day to the reception desk to ask which room is allotted to him.

Question 2.
What was the reaction of Bryson’s wife to his antics?
Answer:
Mrs. Bryson saw the quixotic behaviour of her husband. She was neither angry nor exasperated. But she just expressed her deep sense of wonder. She said, ‘I can’t believe you do this for a living”.

Question 3.
Briefly describe the ‘accidents’ encountered on the flight by Bryson.
Answer:
Once the author was trying to tie a shoe lace. He bent to tie it. Just that moment, the passenger in the front seat reclined in full swing. The author was pinned down in a crash position. On another occasion, he engaged an attractive woman in a witty conversation. While talking to her, he sucked his pen. The pen left a scrub-resistant navy blue on his gum, mouth, teeth and his chin for several days.

C. Based on your understanding of the text, answer the following questions in a paragraph of about 100 – 150 words:

Question 1.
‘To this day, I don’t know how I did it’ – What does ‘it’ refer to?
Answer:
Once, the author knocked a soft drink onto the lap of a sweet little lady sitting beside him. The flight attendant came and cleaned her up and brought him a replacement drink. Instantly he knocked ‘it’ onto the woman again. To this day, the author does not remember how he spilled the drink twice on the same lady passenger seated next to him in the aeroplane.

He just remembers reaching out for the new drink and watch helplessly as his arm, like some cheap prop in one of those 1950 horror movies with a name swept the drink from its perch and on to her lap. The sweet lady was a mm. She looked at him with a stupefied expression. She uttered an oath that started with oh! and ended with sake. In between she used some words the author had never heard before in public.

“I don’t know what is more messy, my room or my life. ”

Question 2.
But, when it’s my own – well, I think hysterics are fully justified’ – How?
Answer:
The author had planned to go to England with all his family members. He arrived at the Logan airport at Boston. When they were checking in, he suddenly remembered that he forgot to use his frequent flier card (British Airways). He also remembered how he had left it in a bag. He tried to open the bag. The zip was jammed. He tried to open it by force. After several attempts, it gave away spilling all the contents in a sprawling corridor in the airport. He ignored the flying documents, silver coins and even passport.

He worried about the tobacco box which was rolling away crazily disgorging its content on the way. He cried “My Tobacco” remembering how expensive it would be to buy tobacco for his pipe in England. Just then he realized that he was bleeding profusely. He had made a gash on his finger while trying to open the zip of his bag by force. He cried hysterically on seeing his own blood, “My finger” My finger”. In general, he was not comfortable flowing other’s blood. But when it came to spilling his own blood “hysterics” was really justified.

“Relived stress through hysterical screaming.”

Question 3.
Bring out the pun in the title ‘The Accidental Tourist’ (one who happens to travel by accident or one who meets with accidents often on his or her trips!).
Answer:
The title “accidental tourist” implies that a man travels a lot and is always confused. He gets into trouble because of his unintentional acts and clumsiness. He does not happen to travel by accident because he should buy a ticket, go to the airport and board the aircraft . wit(i careful plan. But during his travel he does meet with numerable accidents.

The later interpretation is very apt for the author. The story depicts many humorous travel experiences like being pinned in a crash position in his own seat by a fellow passenger, spilling drink on a co-passenger, making his own teeth, gum, chin and tongue scrub-resistant navy blue by his unwise mannerism of sucking the pen, while thinking. The author accidentally gets into trouble often. Hence the pun in the use of “accidental” is pertinent.

“Fill your life with adventures, not things Have stories to tell, not stuff to show.”

Question 4.
Can a clumsy person train himself/herself to overcome short comings? How could this be done?
Answer:
Yes, a clumsy person can train himself to overcome his short comings. People with severe Parkinsons’s disease too learn to hold a spoon and eat with great difficulties. Children with multiple disabilities and nervous problems learn to button up their shirts, tie the shoe laces and even assist friends with similar ailments. There is no difficulty in the world that is insurmountable. Practice makes one perfect. The author’s wife, without rebuking him publicly for his callous clumsiness, trains her children to be supportive of their father Mr. Bryson who always spills drinks or bumps on something or even sits on chewing gum or spilled oil.

As clumsiness is not a welcome behaviour pattern among adults, one must learn how to mend oneself and try to overcome one eccentricity per day with deliberate effort. The author does admit to restrain himself to reduce the laundry bill. But if he extends his efforts even during the presence of his family on long distance air-travels, it would be nice. Cultivating an acceptable behaviour in public places is not an impossible task for any sensible man.

‘‘I am not clumsy. It s just the floor that hates me.
The tables and chairs are bullies
And the wall gets in the way.”

Question 5.
As a fellow passenger of Bill Bryson on the flight, make a diary entry describing his clumsy behaviour during the trip and the inconveniences caused to others as a result of his nervousness.
Answer:
On that fatefulday, I was standing infront of the check in counter just behind a crazyman. He was making odd movement with his hands. It appeared that he was trying to open something. He had yanked the zip of his bag open. It spilled ejecting all the important contents. The funny thing was, the funny man was running behind a tin of tobacco which had already disgorged its content. Then he suddenly cried, “My finger”. May be he had hurt his finger. The funniest side of the whole episode was that he ignored passport, currency, coins and other fluttering travel documents. I had my sincere sympathy with the man but I couldn’t help laughing noiselessly because the fellow was really eccentric.

He did not have a sense of proportion. He gave importance to trifles and ignored the major things lying down fluttering in the corridor of the airport. During his flight he disgorged/spilled the contents of his drink and profusely tendered his apology to the lady next to him. He appeared as real as a buffoon when he appeared sheepish with navy blue ink smeared on his gum, teeth, chin and tongue. It reminded me of a lion faced monkey.

‘‘I have always had a reputation as a Bufoon.”

Additional Questions

Question 6.
Bill Bryson “ached to be suave”. Was he successful in his mission? List his “unsuave ways.
Answer:
Bill Bryson expresses his genuine desire to be “suave”. He would love just once in his life time to rise from the dinner table as if he had experienced an “extremely localized seismic event, get into a car without leaving 14 inch coat outride, wear light-coloured trousers without ever discovering at the end of the day that he had at various times and places sat on chewing gum, ice-cream cough syrup and motor oil. No, Bill Bryson was not successful in his mission. Twice he spilled his drinks on a sweet nun who happened to sit next to him. He tried to show off his wisdom to another attractive lady. As usual, he was sucking his pen.

His shirt, teeth and gum carried the unscrubbable navy blue stain for many days. He always did “liquid mischief’. His clumsy behaviour in the aeroplane made the saintly mm use abusive language. To avoid unsuave ways, he gave up air-travel with his family members. His wife and children supported him yet failed to be refined in manners.
‘‘Heroes, well, they don’t live so long.
But they ’re too suave, and we all admire them.”

Vocabulary

A. Foreign words and phrases .
You have come across the French phrases ‘en famille’ an <famille> and ‘bons mots’ ,ban ma:ts in the lesson. Now look at the following phrases and their meanings.

(a) viva voce – /vaivo vausi /- a spoken examination
(b) sine die -/sina’dAii:/- without a date being fixed
(c) resume -/rezju:mei/- a brief summary
(d) rapport -/rae’pa:(r)/- close relationship with good understanding
(e) bonafide – /bauna faidi/- genuine

B. Refer to the dictionary and find out the meanings of the following foreign words /phrases. Use them in sentences of your own:

1. bon voyage
2. in toto
3. liaison
4. ex gratia
5. en masse
6. en route
7. ad hoc
8. faux pas

1. Bon voyage – Express good wishes to some on leaving for a long journey.Hemalatha went to the airport and said “Bon voyage” to Keerthi who was about to leave for UK.

2. In toto- as a wholeThey accepted the business plan of Murali in toto.

3. Liaison – a close working relationship between people and organization. .The bank clerk regretted his liaison with the watchman who robbed the bank and vanished.

4. Ex gratia-compensation paid by the Government to the victim of an accident.
The Hon’ble Chief Minister gave each of the eight survivors of the Road accident an ex gratia of . two lakh rupees.

5. En masse – in a group all together
The striking workers ran en masse to the gate when the boss arrived.

6. Enroute – on the way
He stopped in Mumbai enroute to Kolkatta.

 

7. Adhoc – created for a particular
An adhoc committee was set up to probe the scam in the universities.

8. Faux pas – an embarrassing or tactless act, blunder
I did not disclose his fauxpas till he joined a college. .

Here is a list of some words borrowed from Indian languages and have been included in the Dictionary of English. Add more words to the table.

WORD	ORIGIN	MEANING
veranda	Hindi	a roofed platform along the outside of a house
bungalow	Hindi	a house in the Bengal style
chutney	Hindi	a ground or mashed relish
cheetah	Sanskrit	uniquely marked
coir	Malayalam	rope
bamboo	Kannada	wood
bandicoot	Telugu	kind of rat
catamaran	Tamil	multi-hulled watercraft
guru	Sanskrit	master

C. Idioms

Look at the list of idioms given below. Find their meanings from a dictionary. Read the sentences that follow and replace the words in italics with the appropriate idioms, making suitable changes wherever necessary.

(a) right up one’s alley –
(b) drive one up the wall –
(c) hit the road-
(d) take (one) for a ride –
(e) in panic mode-

(a) The old man got irritated at the loud noise outside.
(b) We were driving, when it started raining heavily. After stopping for an hour, we began the journey again.
(c) Ramesh gave false excuses for not attending the meeting and deceived me.
(d) At the interview when questions were fired at me rapidly, I forgot every thing and grew irritated.
(e) I love thrillers and this book appeals to me strongly.
Answers
(a) driven up the wall
(b) hit the road
(c) he took me for a ride
(d) got into a panic mode
(e) is right up to my alley

listening Activity

Bala : Hello, Mahesh; where did you spend your holidays?
Mahesh : Well Bala, I had been to a place of ethereal beauty, Kerala.
Bala : Did you spend the three day holiday at Kerala?
Mahesh : Yes Bala. I visited Trivandrum, Quilon and Cochin.
Bala : What are the important tourist spots you visited there?
Mahesh : I shall just mention a few – The Art Museum called Chitralayam at Trivandrum, the King’s palace and the Kovalam Beach.
Bala : What did you see in Quilon?
Mahesh : The lovely scenery at Varkala, the oldest port of Quilon and the Periyar Lake Wild Life Sanctuary.
Bala : Tell me something about Cochin.
Mahesh : Cochin has earned the title ‘Venice of the East’. It is famous for coir and there are many historical monuments there. .
Bala : Oh, I see! I think you should take me to these spots next time you visit Kerala.
Mahesh : Of course! We shall explore its beauties and learn more about our motherland.

Listen to the dialogue read out by the teacher or to the recorded version and answer the questions that follow:

Question (i)
______ was one of the places visited by Mahesh.
(a) Srilanka
(b) Goa
(c) Kasi
(d) Cochin
Answer:
(d) Cochin

Question (ii)
The Art Museum at Trivandrum is called ________
(a) Swamalayam
(b) Gitalayam
(c) Chitralayam
(d) Saranalayam
Answer:
(c) Chitralayam

Question (iii)
Varkala is the oldest port of ________
(a) Quilon
(b) Andhra
(c) Puducherry
(d) the Andamans
Answer:
(a) Quilon

Question (iv)
Mahesh had been to the ________ Lake Wild Life Sanctuary.
(a) Chidambaram
(b) Pulicat
(c) Kovalam
(d) Periyar
Answer:
(d) Periyar

Question (v)
Cochin is called the ________ of the East.
(a) Granary
(b) Cuba
(c) Venice
(d) Pearl
Answer:
(c) Venice

Speaking Activity

(a) Build a dialogue of 8-10 exchanges between your friend and yourself, on the following situation:

You were to board a train to Delhi. By mistake you got into the wrong train and fought for your seat there. On realising your mistake, you left the train shamefaced, after creating a commotion there. Role-play this situation before the class.
James : Last Friday I was caught in an embarrasing situation.
Rahim: Where? How?
James : All the passengers were waiting at the Chennai central station.
Rahim : Where were you heading to? .
James : Well, I was heading to New Delhi.
Rahim: What was the cause of your embarrassment?
James : I did not know about the last minute change of platform. I had got an open ticket to Delhi. Along with many other passengers, I rushed to the unreserved compartment and I boarded it in platform 6. But the platform was changed to 11. I got a comer seat and had a sense of pride.
Rahim : Then when did you realize your mistake?
James : A well-dressed boy asked me if I was going to Mysore. I was surprised and said that I was going to Delhi. He laughed aloud and said to the fellow passengers look, this young man is going to Delhi by Cauvery express. The train had started moving. I rushed out with my luggage and jumped out.
Rahim : Did you catch Tamil Nadu express or not?
James : Of course I did, but I had to travel standing for one night.
Rahim : I’m really sorry to hear it. Be careful in future.

(b) Speak to the class for a minute, as to how one should conduct oneself on formal occasions. (You could talk about table-manners especially while eating, general appearance, manner of speaking, etc.)

Manners maketh a man. While talking, one must look into the eyes of the person spoken to.
When someone else is talking, one should have the decency to pay attention and restate what was gathered to confirm one has understood the essence of the talk. When someone is asking something or enquiring something, one should not be meddling with other things. Such a behaviour amounts to an insult to the speaker. While eating along with friends, when you finish, never get up and leave the table to wash the hands. While eating, one must eat without making much noise. Care must be taken not to spill curry, drinks on the fellow diners.

While attending an interview or going to the college, one must wear neat dress and greet others with a cheerful face. While talking to elders one must always use “Sir” or Madam. While requesting use “please”. When receiving a help, remember to say ‘thank you’. When someone is hurt by an un wise remark, be quick to apologise or say “sorry”.

Reading

Caesar, the Hero of Mumbai on 26/11

1. Mumbai Caesar, the last surviving hero of his kind, died after the attack on one Thursday. Caesar, a Labrador retriever, was covered with tri-colour and given an emotional farewell from the city Police Force. The Mumbai Police Commissioner too marked the passing of the hero with a tweet.

2. Caesar, who was 11 years old was the sole survivor among the dogs of Mumbai Police who took part in bomb detection operations during the terrorist attack on Mumbai that began on November 26, .2008. He died of heart attack at a farm in Virar where he and his three canine buddies had been sent after retirement. During the terror attack in Mumbai, Caesar saved several lives when he sniffed out the hand grenades left by the terrorists at the busy CST railway station.

 

3. Caesar was also a part of the search team at Nariman house, where terrorists were held up for three days. Earlier he was also pressed into service for bomb search operation after the 2006 serial train blasts and July 2017 blast in Mumbai. The Mumbai police officials also tweeted their grief saying, “Services of retired members of Dog Squad during 26/11 will be unforgettable. We will remember our heroes forever.”

Answer the following questions.

Question 1.
Labrador retriever was covered with tri-colour. What does this signify?
The use of tri colour flag on the body of Labrador retriever signifies that it has served the nation like a soldier and deserves our homage.

Question 2.
How did Caesar save several lives at the CST railway station?
Answer:
During the terror attack at Mumbai, Caesar saved several lives when he sniffed out the hand grenades left by the terrorists at the busy CST railway station.

Question 3.
Which word in the passage (paragraph 3) means the same as ‘forced’.
Answer:
Pressed into means “forced”.

Question 4.
“Services of retired members of Dog Squad during 26/11 will be unforgettable”. Mention three services rendered by Caesar.
Answer:
Caesar had taken part in the bomb detection operations during the terrorist attack on Mumbai that began on Novmeber 26, 2008. Caeser saved several lives when he sniffed out the hand grenades left by the terrorists at the busy CST railway station. Caesar was also a part of bomb search operation after 2006 serial train blasts and July 2017 blast in Mumbai.

Question 5.
Caesar is a Labrador breed of dogs. Name a few other native breeds that are used by the Police force.
Answer:
The Blood hound, The German short haired pointer, The Boxer, The Doberman, The Dutch German Shepherd and The Giant Schnauzer are some of the famous breeds used by police force.

Question 6.
Try to rewrite the news item in your mother tongue without losing the spirit and flavour of the text. Give a suitable title to your translated version.
Answer:
வீரமரணம்
மும்பை 26\11 தாக்குதலில் குண்டு கண்டுபிடிக்கும் வீரநாய் சீசர் மரணம்.
1. மும்பை சீசர், அவரது சகாக்களில் கடைசியாக எஞ்சியிந்த மும்பை தாக்குதலுக்குப் பின் ஓர் வியாழக்கிழமை மரணமடைந்தது மீட்புப் பணியில் ஈடுபட்ட அந்த லப்ராடர் இனநாய் மூவ்வண்ணக் கொடியால் போர்த்தப்பட்டு ஓர் உணர்ச்சி மயமான பிரிவு உபச்சாரம் நகர காவல்துறையால் வழங்கப்பட்டது. சீசரின் வீரமரணத்தைப் பற்றி மும்பை காவல் ஆணையர் ட்வீட் செய்துள்ளார்.

2. பதினோறு வயது நிரம்பிய சீசர் மும்பை காவல்துறையில் பணியில் இருந்தது. தீவிரவாதத் தாக்குதலின் போது குண்டு கண்டுபிடிக்கும் பணியில் நவம்பர் 26, 2008 முதல் ஈடுபடுத்தப்பட்டு பயிற்சி பெற்ற பல நாய்களில் உயிர்தப்பிய ஒரே ஒரு நாயாகும். விரார் அருகே உள்ள பண்ணை வீட்டில் அது மாரடைப்பால் மரணமடைந்தது. அங்குதான் அதுவும் அதன் மூன்று தோழர்களும் பணி ஓய்வுக்குப் பின் அனுப்பப் பட்டிருந்தனர். மும்பை தீவிரவாதத் தாக்குதலின் போது, சீசர் தீவிரவாதிகள் விட்டுச் சென்ற பல கையெறிகுண்டுகளை மோப்பம் பிடித்து கண்டுபிடித்து மிகவும் பரப்பரப்பாக இயங்கிக் கொண்டிருந்த சத்ரபதிசிவாஜி இரயில் நிறுத்தத்தில் (CST) பல உயிர்களைக் காப்பாற்றியது.

3. மூன்று நாட்களாக தீவிரவாதிகள் பதுங்கியிருந்த நரிமன் பாயின்டில் தேடும் குழுவின் ஓர் அங்கமாக சீசரும் இருந்தது. 2006 ஆம் ஆண்டு தொடர்வண்டி குண்டுவெடிப்புச் சம்பவத்தின் போதும், ஜுலை 2017 மும்பை குண்டுவெடிப்புக்கு முன்னரும் பல முறை குண்டு தேடும் பணியில் அது ஈடுபடுத்தப்பட்டது. மும்பை காவல் அதிகாரிகளும் தமது துயரத்தை ட்விட்டர் வலை தளத்தில் “26/11 நாய் படையிலிருந்து ஓய்வுபெற்றவர்களது சேவை மறக்க இயலாதது. எங்களது வீரர்களை நாங்கள் என்றும் நினைவில் வைத்துப் போற்றுவோம்” எனப் பதிவு செய்தனர்.

Grammar

Now complete the following.

(a) Do as directed.

Question 1.
Dinesh and Prabhu wanted to meet Varsha at the bus stop. They went to the bus stop. (Change into a compound sentence)
Answer:
Dinesh and Prabhu wanted to meet Varsha at the bus stop and so they went to the bus stop.

Question 2.
Varsha reached the railw ay station. She was waiting for them there.(Change into a compound sentence)
Answer:
Varsha reached the railway station and she was waiting for them there.

 

Question 3.
While she waited at the train station, Varsha realized that the train was late. (Change into a simple sentence)
Answer:
Waiting at the railway station, Varsha realized that her train was late.

Question 4.
Dinesh and Prabhu left the bus stop. Varsha rang them. (Change into a complex sentence)
Answer:
After /When Dinesh and Prabhu left the bus stop, Varsha rang them.

Question 5.
The trio met at the station. Varsha left for Madurai. (Change into a complex sentence)
Answer:
After the trio met at the railway station, Varsha left for Madurai.

Question (b)
This paragraph has only simple sentences. Combine them into compound and complex sentences. The first one is done for you.
Answer:
One day Ajay and Tijo went to the canal. They wanted to catch some fish. Some people were playing nearby. They chose a better place. They took out the fishing rods. Suddenly there was a loud splash. They also heard a loud scream. Both Ajay and Tijo looked up. They saw something moving in the water. Then they saw a hand waving. Someone had fallen in the water. It was Yusuf. He had jumped into the water. He wanted to swim.
One day Ajay and Tijo went to the canal to catch some fish

One day Ajay and Tijo wanted to catch some fish and so they went to the canal. As some people were playing nearby, they chose a better place. When they took out the fishing rods, there was a loud splash and scream. As they looked up, they saw something like a hand waving, in the water. It was obvious that someone had fallen into the water. It was Yusuf who had jumped into the water to swim.

Question (c)
Here is one long sentence. Split them into smaller sentences.
Like all living things, human beings also need food in order to live as every part of the body must get a steady supply of food so that it can work properly, but first the food eaten has to be broken down through a process called digestion so that it can dissolve in the blood and carried to all parts of the body.
Answer:
Like all living things, human beings need food to live. Every part of the body must get a steady ‘ supply of food. Only then it can work properly. First the eaten food has to be broken down through digestion. The digested food dissolves into the blood. It is then carried to all parts of the body.

Writing

Now write a short story to explain these proverbs.

Question 1.
Actions speak louder than words.
Answer:
Vivek Pradhan was not a happy man.. Even the plush comfort of the air-conditioned compartment of the Shatabdi express could not cool his frayed nerves. He was the Project Manager and still not entitled to air travel. It was not the prestige he sought, he had tried to reason with the admin person, it was the savings in time. As Project Manager, he had so many things to do!!
He opened his case and took out the laptop, determined to put the time to some good use.

‘Are you from the software industry sir?’ the man beside him was staring appreciatively at the laptop. Vivek glanced briefly and mumbled in affirmation, handling the laptop now with exaggerated care and importance as if it were an expensive car.

‘You people have brought so much advancement to the country, Sir. Today everything is getting computerized.’ ‘Thanks,’ smiled Vivek, turning around to give the man a look. He always found it difficult to resist appreciation. The man was young and stockily built like a sportsman. He looked simple and strangely out of place in that little lap of luxury like a small town boy in a prep school. He probably was a railway sportsman making the most of his free travelling pass.

‘You people always amaze me,’ the man continued, ‘You sit in an office and write something on a computer and it does so many big things outside.’ Vivek smiled deprecatingly. Naiveness demanded reasoning not anger. ‘It is not as simple as that my friend. It is not just a question of writing a few lines. There is a lot of process that goes behind it.’ For a moment, he was tempted to explain the entire Software Development Lifecycle but restrained himself to a single statement.

‘It is complex, very complex.’ ‘It has to be. No wonder you people are so highly paid,’ came the reply. This was not turning out as Vivek had thought. A hint of belligerence crept into his so far affable, persuasive tone. ‘Everyone just sees the money. No one sees the amount of hard work we have to put in. Indians have such a narrow concept pf hard work. Just because we sit in an air-conditioned office, does not mean our brows do not sweat. You exercise the muscle;
‘we exercise the mind and believe me that is no less taxing.’

He could see, he had the man where he wanted, and it was time to drive home the point.

‘Let me give you an example. Take this train. The entire railway reservation system is computerized. You can book a train ticket between any two stations from any of the hundreds of computerized booking centers across the country. Thousands of transactions accessing a single database, at a time concurrently; data integrity, locking, data security. Do you understand the complexity in designing and coding such a system?’ The man was awestruck; quite like a child at a planetarium. This was something big and beyond his imagination.

‘You design and code such things?’ ‘I used to,’ Vivek paused for effect, ‘but now I am the Project Manager.’ ‘Oh! ’ sighed the man, as if the storm had passed over, ‘so your life is easy now.’ This was like the last straw for Vivek. He retorted, ‘Oh come on, does life ever get easy as you go up the ladder. Responsibility only brings more work. Design and coding! That is i the easier part. Now I do not do it, but I am responsible for it and believe me, that is far more

stressful. My job is to get the work done in time and with the highest quality. To tell you about the pressures, there is the customer at one end, .always changing his requirements, the user at the other, wanting something else, and your boss, always expecting you to have finished it i yesterday.’ Vivek paused in his diatribe, his belligerence fading with self-realization. What he had said, was not merely the outburst of a wronged man, it was the truth. And one need not get angry while defending the truth. ‘My friend,’ he concluded triumphantly, ‘you don’t know what it is to be in the Line of Fire’. The man sat back in his chair, his eyes closed as if in realization. When he spoke after sometime, it was with a calm certainty that surprised Vivek. ‘I know sir, I know what it is to be in the Line of Fire ’He was staring blankly, as if no passenger,
no train existed, just a vast expanse of time.

‘There were 30 of us when we were ordered to capture Point 4875 in the cover of the night. The enemy was firing from the top. There was no knowing where the next bullet was going to come from and for whom. In the morning when we finally hoisted the tri-colour at the top only four of us were alive.’ ‘You are a…?’ ‘I am Subedar Sushant from the 13 J&K Rifles on duty at Peak 4875 in Kargil. They tell me 1 have completed my term and can opt for a soft assignment. But, tell me sir, can one give up duty just because it makes life easier? On the dawn of that capture, one of my colleagues lay injured in the snow, open to enemy fire while we were hiding behind a bunker.

It was my job to go and fetch that soldier to safety. But my captain sahib refused me permission and went ahead himself. He said that the first pledge he had taken as a Gentleman Cadet was to put the safety and welfare of the nation foremost followed by the safety and welfare of the men he commanded his own personal safety came last, always and every time.’

‘He was killed as he shielded and brought that injured soldier into the bunker. Every morning thereafter, as we stood guard, I could see him taking all those bullets, which were actually meant for me . I know sir….I know, what it is to be in the Line of Fire.’ Vivek looked at him in disbelief not sure of how to respond. Abruptly, he switched off the laptop. It seemed trivial, even insulting to edit a Word document in the presence of a man for whom valour and duty was a daily part of life; valour and sense of duty which he had so far attributed only to epical heroes. The train slowed down as it pulled into the station, and Subedar Sushant picked up his bags to alight.

‘It was nice meeting you sir.’ Vivek fumbled with the handshake.

This hand… had climbed mountains, pressed the trigger, and hoisted the tri-colour. Suddenly, as if by impulse, he stood up in attention and his right hand went up in an impromptu salute,… It was the least he felt he could do for the country.

PS: The incident he narrated during the capture of Peak 4875 is a true-life incident during the Kargil war. Capt. Batra sacrificed his life while trying to save one of the men he commanded, as victory was within sight. For this and various other acts of bravery, he was awarded the Param Vir Chakra, the nation’s highest military award. Live humbly, there are great people around us, let us learn! Action speaks louder than words

 

Question 2.
Despair gives courage to a coward.
Answer:
Most of the time, we do not realize the talent that is within us. Certain situations help to bring out these special qualities to the fore. Just as a drowning man tries to hold on to anything that comes his way to save his life, we too come up with remedies beyond our imagination in desperate situations. We find the courage to do extraordinary things when we are in despair. Even a coward can be turned into a brave warrior in a life-threatening situation. Once there was an ant that lived in a tree by the side of a pond. He was very timid and was scared of even the slightest disturbance in the tree. One day a heavy wind blew across the tree and the ant fell into the pond and was struggling to swim and stay afloat. A pigeon that lived in the same tree broke a twig of the tree and threw it to the ant. The ant got on to the twig and the pigeon swooped down picked up the twig and placed it back on the tree. The ant thanked the pigeon profusely for saving his life.

As the days went by the ant and the pigeon became good friends. The pigeon always told the ant that he should learn to be a bit braver and face life boldly. But the ant remained the same. One day a hunter came to the pond to quench his thirst. As he was drinking water from the pond he noticed the pigeon sitting on the tree. The hunter wanted to shoot the pigeon and got his bow and arrow ready. The ant noticed this and wanted to desperately save his friend. The pigeon was fast asleep and there was no way to warn him since he was sitting a long way the ant jumped down from the tree and crawled as fast as he could and bit the toe of the hunter hard as he could. The hunter cried out in pain and missed his aim.

The pigeon woke up hearing the noise and noticed the hunter with his bow and arrow. It flew away to safety. The ant was very happy that it could save his friend and felt elated for having acted bravely for once in his life.

The story clearly illustrates that when placed in a desperate situation the ant could rise above its limitations and act bravely in order to save the pigeon’s life. So when the situation demands even a coward can turn into a courageous person.

Develop the following hints into a paragraph

As is the King, So are the Subjects

Once upon a time two kings ruled neighbouring kingdoms. King Arya was a great warrior. He looked after his subjects very well. People loved him. He was always looking for ways to increase their safety and welfare. All his subjects were happy. On the other hand, king Vaishal was a very lazy man. He spent his time entertaining himself. There was always singing, dancing and merry making. His subjects were very angry with him as he never came out of the palace to listen to their woes.

A powerful Sultan attacked both the kingdoms. King Arya’s army, being well-prepared was

Very alert. The enemy forces were really powerful. But men, women and even children i joined the hands of army to protect their king. King Vaishal’s subject soon after sensing the

impending war, started fleeing the kingdom. King was left all alone. People were not interested in protecting him as he was not interested in their welfare and safety.

King Vaishal realized his foolishness but it was too late. He was defeated in war and fled his country in disgrace to save his own life. King Arya defeated Sultan. This happened because King Arya always kept the welfare of his people in his heart all the time. His subjects reciprocated his love. They were loyal and supportive during testing times.

Writing A Curriculum Vitae

Vijayraj Joseph
Task: Write a CV for the post of a DTP operator at ABC Publishing House and send it to P.O. Box No. 2345 or E-mail it to abcph@nomail.com

Curriculum Vitae
Name : Vijay
Mobile No : 8765412385
email : vijayl975@gmail.com
Address : 17/2 Beach road,
Neelangarai Chennai

Career objective:
Looking for a challenging job in the field of Desk top publishing-which requires the optimum use of my skills in typing and designing and provides me opportunities for vertical growth.

Synopsis:
A graduate with a degree in computer science from Loyola college, Madras

Profile

• Good working knowledge of computers.
• Good at typing .
• Excellent at wrapper designing and editing
• Profound knowledge in Corel draw and photoshop
• Expertise in MS word and Excel file handling.

Educational Qualification

• B.Sc Computer Science with a second class
• 12th with aggregate of 67% from GHSS, Koovathur
• 10th with aggregate of 55% from GHSS, pudhupattinam, Kalpakkam

Previous Experience

• Two years experience in student Xerox as a DTP operator

Projects done

• Typed about 50 PhD dissertation and 20 M. Phil thesis

Extra-curricular Activities

• NSS volunteer in college
• Volleyball District player Strength:
• A team player
• Devoted and smart in work
• Optimistic

Personal Details:
Date of birth : 06.09.1993
Sex : Male
Marital status : Single
Languages known : Tamil and English

Declaration
I hereby declare that the above furnished information is true to the best of my knowledge. If I am offered an opportunity, I shall prove my mettle and be worthy of your choice
S/d
Vijay

Task 1:
You see an advertisement in the newspaper. A publishing house in Chennai has brought out a paperback edition of the complete works of Khushwant Singh. You want to buy it. You are asked to send a Demand Draft for Rs.1000/- Fill in the following DD challan in favour of ‘X publishing house, New Delhi’, payable at Chennai. The surcharge for Rs.1000/- is Rs.25/-

Answer:

Task 2:
Fill in the following forms with imaginary details.
Question (a)

Answer:

Question (b)

Answer:

Question (c)

Answer:

The Accidental Tourist About the Author

Bill Bryson is an Anglo-American author of books on travel, English language, science and non-fiction topics. He stayed in Great Britain during his adult years. He served as the Chancellor of Durham University from 2005 to 2011. ’Neither here not there’, ‘Notes from a small island’, ‘A Walk in the woods’
‘A Short history of nearly everything’, ‘The Life and Times of the Thunderbolt Kid’, and Icons of England are some his famous

The Accidental Tourist Summary

The whole story revolves around the author’s inability to do ordinary things that other people do in public places like a cinema theatre or a plane. He has problems of orientation. He would try to reach a rest room but find himself behind a self-locking door. The author recounts how his anxiety ruined the happiness of people at the airport and how he himself suffered. For fear of such life-time incompatibility, the author refrained from making frequent trips and could never use his frequent flier miles for a trip to Bali island. The author had a flash of memory that he had joined British Airways frequent flier programme. His entire family arrived at Logan Airport in Boston. Unfortunately, he had placed the card in a carry-on-bag hanging around his neck.

The zip on’ the bag was jammed. He tried pulling it hard. He used force instead of commonsense, the zip gave way spilling the contents over an area as large as tennis court.

To the great embarrassment of all fellow passengers waiting at the airport, 14 ounce tin pipe, coins and other important documents flew helter-skelter. He was annoyed to see the snuff box rolling and emptying its content. Instead of trying to collect coins, documerit and important papers, he went behind the tobacco box crazily shouting, “My tobacco”. When he sensed oozing of blood from his finger that he had gashed while opening the bag’s jammed zip he shifted his attention immediately to his finger. The author’s wife, wondering at his hysterics said, “I can’t believe you do this for a living”.

Once in an aeroplane, the author leaned to tie his shoe lace. Just then some one in the front seat leaned back into full recline. The author found himself pinned helplessly in a crash position. Once the author spilled drinks on a fellow passenger too. He did it twice. He did not know how he did it.

The author was recording his important thoughts in a note book. They were as silly as he was “Buy socks, clutch drinks carefully”. He had fallen into a deep conversation with an attractive young lady in the next seat. Having lost his head, he amused her for about twenty minutes exchanging witty remarks to impress her. When he went to a rest room, he realized that the pen head leaked. His mouth, teeth and gums appeared in striking, scrub-resistent navy blue. To his disgust, it remained blue for many days.

The author wanted to be polished and sophisticated in his manners; But as serendipity would have it, he never left a dinner without making fellow diners realize that he had just experienced a local seismic event. Whenever he entered a car, he closed the door when his coat was still 14 inches outside the door. His trousers always brought evidence of his having sat on chewing gum, ice-cream, cough syrup or motor oil.

The author had clear prediction of the catastrophe he intended to venture into unconsciously. So, his wife would give directions to children like,

“Take the lid off the food for daddy” or put your hoods up children, Daddy’s going to cut his meat”. The author does all such cranky things only during his flight with his family.

While staying alone the author sat on his own down turned palms to prevent himself from doing some “liquid mischief ”. When alone, he never does things which would land him in some sure catastrophe.

The author admits his peculiar inability of earning flyer miles. He has a habitual forgetting of his “Frequent flyer mile card” while collecting his Boarding pass. Though he deserves to have earned 100,000 miles a year, and visited Bali on a free air ticket, he had earned only 212 air miles. While on a flight to Australia, the airline agency refused to grant air miles because the flyer mile card read as W. Bryson where as the name in the ticket read as B. Bryson.

 

He admits that he would never go to Bali because he has a peculiar habit of spilling the contents of food on board. People who have clumsy habits do overcome their awkwardness in public places with careful practice. As we live in a society, we should leam to conform to the norms of the society including table manners.

Textual:
alley – a narrow passage-way between or behind buildings
Bill – William (the letter W is changed to B and William is called Bill)
bons mots – (French) witty remarks
cascade – waterfall
catastrophe – a terrible disaster
concourse – the open central area in a
large public building (here‘airport’) /hall
consternation – worry
disgorging -discharging
en famille – (French) as a family
exasperation – irritation
extravagantly – excessively
gashed – cut deeply
hysterics – a fit of uncontrollable laughing or crying
suave – polite and sophisticated
venerable – valued
yanked – pulled with a jerk

Additional:
abruptly – suddenly
accumulated – collected
annoyed – angry and irritated
anxiety – worry
confused – disoriented
constantly – all the time
dumbstruck – shocked and speechless
entitled – deserving special treatment
evident – obvious
frequent – often
frustration – vexation
hysteries – uncontrolled emotion
jammed – packed tightly
oblivions – forgetfulness
panic – intense fear
specialty – particular skill

The Accidental Tourist Synonyms

Choose the most appropriate synonyms for the underlined words,

Question 1.
I am constantly filled with wonder.
(a) hardly
(b) regularly
(c) rarely
(d) sparsely
Answer:
(b) regularly

Question 2.
I yanked at the zip of the bag.
(a) pushed
(b) dragged
(c) jerked
(d) closed
Answer:
(c) jerked

Question 3.
He had always ended up in standing in an alley.
(a) maze
(b) road
(c) pathway
(d) hospital
Answer:
(c) pathway

Question 4.
He tried pulling it with great consternation.
(a) anger
(b) hatred
(c) pity
(d) worry
Answer:
(d) worry

 

Question 5.
All the contents of the bag were extravagantly ejected.
(a) frugally
(b) judiciously
(c) vainly
(d) lavishly
Answer:
(d) lavishly

Question 6.
The tin of tobacco rolled crazily across the concourse,
(a) closure
(b) locked up
(c) hall
(d) strait
Answer:
(c) hall

Question 7.
I had gashed my finger.
(a) bandaged
(b) plastered
(c) cut
(d) healed
Answer:
(c) cut

Question 8.
The tobacco tin went disgorging its content.
(a) attacking
(b) discharging
(c) collecting
(d) dusting
Answer:
(b) discharging

Question 9.
I always have catastrophies when I travel.
(a) condy
(b) disasters
(c) jokes
(d) joy
Answer:
(b) disasters

Question 10.
The author’s wife did not show anger or exasperation.
(a) pleasure
(b) irritation
(c) pain
(d) hatred
Answer:
(b) irritation

Question 11.
He amused her with urbane bons mots.
(a) news
(b) puzzles
(c) witticism/repartee jokes
(d) joked
Answer:
(c) witticism/repartee jokes

Question 12.
He explained the venerable relationship between Bill and William but in vain.
(a) loathsome
(b) valued
(c) hurt
(d) howled
Answer:
(b) valued

Question 13.
I ache to be suave.
(a) rude
(b) indecent
(c) dishonest
(d) polite / sophisticated
Answer:
(d) polite / sophisticated

 

Question 14.
When it’s my own blood, I think hysterics is justified.
(a) frenzied
(b) whisper
(c) rustle
(d) hustle
Answer:
(a) frenzied

Question 15.
My hair went into panic mode.
(a) courage
(b) fear
(c) joy
(d) pleasure
Answer:
(b) fear

The Accidental Tourist Antonyms

Choose the most appropriate antonyms for the underlined words.

Question 1.
I was shedding blood in a lavish manner.
(a) Extravagant
(b) frugal
(c) easy
(d) tough
Answer:
(b) frugal

Question 2.
I am constantly filled with wonder.
(a) inconstantly/rarely
(b) incorrectly
(c) infrequently
(d) incessantly
Answer:
(a) inconstantly/rarely

Question 3.
I pulled it with a frown.
(a) Scowl
(b) grimace
(c) mockery
(d) smile
Answer:
(d) smile

Question 4.
I managed to get myself freed.
(a) cleared
(b) disentagled
(c) imprisoned trapped
(d) discharged
Answer:
(c) imprisoned trapped

Question 5.
I watched dumbstruck as the carefully sorted documents came down in a cascade.
(a) ruffled
(b) excited
(c) expected
(d) petrified
Answer:
(c) expected

Question 6.
I was talking to an attractive lady.
(a) charming
(b) captivating
(c) unattractive / ugly
(d) ravishing
Answer:
(c) unattractive / ugly

Question 7.
I cried in horror.
(a) alarm
(b) antipathy
(c) disgust
(d) pleasure
Answer:
(d) pleasure

Question 8.
The author amused the lady with some witty remarks.
(a) bored
(b) doted
(c) delighted
(d) entertained
Answer:
(a) bored

Question 9.
Coins bounced to variety of noisy oblivions.
(a) Nuisance
(b) awareness
(c) forgetfulness
(d) unconscious
Answer:
(b) awareness

 

Question 10.
This had become a real frustration.
(a) Disappointment
(b) cramp
(c) discontentment
(d) fulfilment
Answer:
(d) fulfilment

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Prose Chapter 6 The Accidental Tourist Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

For those looking for help on 11th Physics can use the Tamilnadu State Board Solutions for 11th Physics Chapter 11 Waves prevailing for free of cost.

Download the Samacheer Kalvi 11th Physics Book Solutions Questions and Answers Notes Pdf, for all Chapter Wise through the direct links available and take your preparation to the next level. The Tamilnadu State Board Solutions for 11th Physics Chapter 11 Waves Questions and Answers covers all the topics and subtopics within it. Practice using these Samacheer Kalvi 11th Physics Book Solutions Questions and Answers for Chapter 11 Waves PDF and test your preparation level and bridge the knowledge gap accordingly.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

If you have any queries take the help of the Tamilnadu State Board Solutions for 11th Physics Chapter 11 Waves Questions and Answers learn all the topics in it effectively. We have everything covered and you can Practice them often to score better grades in your exam. By going through the Samacheer Kalvi 11th Physics Book Solutions Questions and Answers you can attempt the actual exam with utmost confidence.

Samacheer Kalvi 11th Physics Waves Textual Evaluation Solved

Samacheer Kalvi 11th Physics Waves Multiple Choice Questions
Question 1.
A student tunes his guitar by striking a 120 Hertz with a tuning fork, and simultaneously plays the 4^th string on his guitar. By keen observation, he hears the amplitude of the combined sound oscillating thrice per second. Which of the following frequencies is the most likely the frequency of the 4^th string on his guitar?
(a) 130
(b) 117
(c) 110
(d) 120
Answer:
(b) 117

Question 2.
A transverse wave moves from a medium A to a medium B. In medium A, the velocity of the transverse wave is 500 ms^-1 and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B when its velocity is 600 ms^-1, respectively are
(a) 120 Hz and 5 m
(b) 100 Hz and 5 m
(c) 120 Hz and 6 m
(d) 100 Hz and 6 m
Answer:
(d) 100 Hz and 6 m

Question 3.
For a particular tube, among six harmonic frequencies below 1000 Hz, only four harmonic frequencies are given : 300 Hz, 600 Hz, 750 Hz and 900 Hz. What are the two other frequencies missing from this list?
(a) 100 Hz, 150 Hz
(b) 150 Hz, 450 Hz
(c) 450 Hz, 700 Hz
(d) 700 Hz, 800 Hz
Answer:
(b) 150 Hz, 450 Hz
Hint:
If the tube is open at both ends so the harmonic frequencies are based on 150 Hz.
1^st = 150 Hz ; 2^nd = 300 Hz ; 3^rd = 450 Hz ; 4^th = 600 Hz ; 5^th = 750 Hz ; 6^th = 900 Hz
The above frequencies the missing frequency in the list 150 Hz, 450 Hz

Question 4.
Which of the following options is correct?

Options for (1), (2) and (3), respectively are
(a) (B), (C) and (A)
(b) (C), (A) and (B)
(c) (A), (B) and (C)
(d) (B), (A) and (C)
Answer:
(a) (B), (C) and (A)

Question 5.
Compare the velocities of the wave forms given below, and choose the correct option.

where, v_A, v_B, v_C and v_D are velocities given in (A), (B), (C) and (D), respectively.
(a) V_A > V_B > V_D > V_C
(b) V_A < V_B < V_D < V_C
(c) V_A = V_B = V_D = V_C
(d) V_A > V_B = V_D > V_C
Answer:
(c) V_A = V_B = V_D = V_C

Question 6.
A sound wave whose frequency is 5000 Hz travels in air and then hits the water surface. The ratio of its wavelengths in water and air is …….
(a) 4.30
(b) 0.23
(c) 5.30
(d) 1.23
Answer:
(a) 4.30
Hint.
Frequency of sound, f = 5000 Hz

Question 7.
A person standing between two parallel hills fires a gun and hears the first echo after t₁ sec and the second echo after t₂ sec. The distance between the two hills is …..

Answer:

Hint:

Question 8.
An air column in a pipe which is closed at one end, will be in resonance with the vibrating body of frequency 83 Hz. Then the length of the air column is ………
(a) 1.5 m
(b) 0.5 m
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m
Hint:

Velocity of sound in air y = 343 ms^-1

Question 9.
The displacement y of a wave travelling in the x direction is given by where x and y are measured in metres and t in second. The speed of the wave is ………
(a) 150 ms^-1
(b) 300 ms^-1
(c) 450 ms^-1
(d) 600 ms^-1
Answer:
(a) 150 ms^-1
Hint:

Question 10.
Consider two uniform wires vibrating simultaneously in their fundamental notes. The tensions, densities, lengths and diameter of the two wires are in the ratio 8 : 1, 1 : 2, x : y and 4 : 1 respectively. If the note of the higher pitch has a frequency of 360 Hz and the number of beats produced per second is 10, then the value of x : y is ……….
(a) 36 : 35
(b) 35 : 36
(c) 1 : 1
(d) 1 : 2
Answer:
(a) 36 : 35

Question 11.
Which of the following represents a wave?
(a) (x – vt)³
(b) x(x + vt)
(c) \(\frac{1}{(x+v t)}\)
(d) sin (x + vt)
Answer:
(d) sin (x + vt)

Question 12.
A man sitting on a swing which is moving to an angle of 60° from the vertical is blowing a whistle which has a frequency of 2.0 k Hz. The whistle is 2.0 m from the fixed support point of the swing. A sound detector which detects the whistle sound is kept in front of the swing. The maximum frequency the sound detector detected is …….
(a) 2.027 kHz
(b) 1.947 kHz
(c) 9.74 kHz
(d) 1.011 kHz
Answer:
(a) 2.027 kHz

Question 13.
Let. y = \(\frac{1}{1+x^{2}}\) at t = 0s be the amplitude of the wave propagating in the positive x-direction. At t = 2s, the amplitude of the wave propagating becomes . Assume that the shape of the wave does not change during propagation. The velocity of the wave is …..
(a) 0.5 ms^-1
(b) 1.0 ms^-1
(c) 1.5 ms^-1
(d) 2.0 ms^-1
Answer:
(b) 1.0 ms^-1
Hint.
The general expression y in terms of x

The shape of wave does not change, also wave move in 2 sec, 2m in positive ‘x’ direction. So, wave moves 2m in 2 sec.

Question 14.
A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. Which of the following plots shows the correct variation of speed v with height h from the lower end?

Answer:

Question 15.
An organ pipe A closed at one end is allowed to vibrate in its first harmonic and another pipe B open at both ends is allowed to vibrate in its third harmonic. Both A and B are in resonance with a given tuning fork. The ratio of the length of A and B is …….

Answer:
(c) \(\frac{1}{6}\)
Hint:
(c) \(\frac{1}{6}\)

Samacheer Kalvi 11th Physics Waves Short Answer Questions

Question 1.
What is meant by waves?
Answer:
The disturbance which carries energy and momentum from one point in space to another point in space without the transfer of the medium is known as a wave.

Question 2.
Write down the types of waves.
Answer:
Waves can be classified into two types:
(a) Transverse waves
(b) Longitudinal waves

Question 3.
What are transverse waves? Give one example.
Answer:
In transverse wave motion, the constituents of the medium oscillate or vibrate about their mean positions in a direction perpendicular to the direction of propagation (direction of energy transfer) of waves.
Example: light (electromagnetic waves)

Question 4.
What are longitudinal waves? Give one example.
Answer:
In longitudinal wave motion, the constituent of the medium oscillate or vibrate about their mean positions in a direction parallel to the direction of propagation (direction of energy transfer) of waves.
Example: Sound waves travelling in air.

Question 5.
Define wavelength.
Answer:
For transverse waves, the distance between two neighbouring crests or troughs is known as the wavelength. For longitudinal waves, the distance between two neighbouring compressions or rarefactions is known as the wavelength. The SI unit of wavelength is meter.

Question 6.
Write down the relation between frequency, wavelength and velocity of a wave.
Answer:

Question 7.
What is meant by interference of waves?
Answer:
Interference is a phenomenon in which two waves superimpose to form a resultant wave of greater, lower or the same amplitude.

Question 8.
Explain the beat phenomenon.
Answer:
When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two. sources, then their difference in frequency gives the beat frequency. Number of beats per second n = | f₁ – f₂| per second

Question 9.
Define intensity of sound and loudness of sound.
Answer:

1. The loudness of sound is defined as “the degree of sensation of sound produced in the ear or the perception of sound by the listener”.
2. The intensity of sound is defined as “the sound power transmitted per unit area taken normal to the propagation of the sound wave”.

Question 10.
Explain Doppler Effect.
Answer:
When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.

Question 11.
Explain red shift and blue shift in Doppler Effect.
Answer:
If the spectral lines of the star are found to shift towards red end of the spectrum (called as red shift) then the star is receding away from the Earth. Similarly, if the spectral lines of the star are found to shift towards the blue end of the spectrum (called as blue shift) then the star is approaching Earth.

Question 12.
What is meant by end correction in resonance air column apparatus?
Answer:
The antinodes are not exactly formed at the open end, we have to include a correction, called end correction e, by assuming that the antinode is formed at some small distance above the open end. Including this end correction, the first resonance is

Again taking end correction into account, we have

Question 13.
Sketch the function Y = x + a. Explain your sketch
Answer:
When a = 0, y = x
when a = 1; x = 1: y = 1 + 1 = 2
when a = 2; x = 2; y = 2 + 2 = 4

Explanation: This implies, when increasing the value of a, the line shifts towards right side at a = 0, and line shifts towards left side at a = 1, 2, ….For a = vt, y = x – vt satisfies the differential equation. Though this function satisfies the differential equation, it is not finite for all values of x and t. Hence it does not represent a waves.

Question 14.
Write down the factors affecting velocity of sound in gases.
Answer:
(a) Effect of pressure
(b) Effect of temperature
(c) Effect of density
(e) Effect of wind

Question 15.
What is meant by an echo? Explain.
Answer:
Echo: An echo is a repetition of sound produced by the reflection of sound waves from a wall, mountain or other obstructing surfaces.
Explanation: The speed of sound in air at 20°C is 344 m s^-1. If we shout at a wall which is at 344 m away, then the sound will take 1 second to reach the wall. After reflection, the sound will take one more second to reach us. Therefore, we hear the echo after two seconds. Scientists have estimated that we can hear two sounds properly if the time gap or time interval between each sound is \(\left(\frac{1}{10}\right)^{\text {th }}\) of a second (persistence of hearing) i.e., 0.1 s. Then,

2d= 344 × 0.1 = 34.4m ;d= 17.2m
The minimum distance from a sound reflecting wall to hear an echo at 20°C is 17.2 meter.

Samacheer Kalvi 11th Physics Waves Long Answer Questions

Question 1.
Discuss how ripples are formed in still water.
Answer:
Suppose we drop a stone in a trough of still water, we can see a disturbance produced at the place where the stone strikes the water surface. We find that this disturbance spreads out (diverges out) in the form of concentric circles of ever increasing radii (ripples) and strike the boundary of the trough. This is because some of the kinetic energy of the stone is transmitted to the water molecules on the surface. Actually the particles of the water (medium) themselves do not move outward with the disturbance. This can be observed by keeping a paper strip on the water surface. The strip moves up and down when the disturbance (wave) passes on the water surface. This shows that the water molecules only undergo vibratory motion about their – mean positions.

Question 2.
Briefly explain the difference between travelling waves and standing waves.
Answer:

Question 3.
Show that the velocity of a travelling wave produced in a string is v =\(\sqrt{\frac{\mathrm{T}}{\mu}}\)
Answer:
Velocity of transverse waves in a stretched string: Let us compute the velocity of transverse travelling waves on a string. When a jerk is given at one end (left end) of the rope, the wave pulses move towards right end with a velocity v. This means that the pulses move with a velocity v with respect to an observer who is at rest frame. Suppose an observer also moves with same velocity v in the direction of motion of the wave pulse, then that observer will notice that the wave pulse is stationary and the rope is moving with pulse with the same velocity v. Consider an elemental segment in the string. Let A and B be two points on the string at an instant of time. Let dl and dm be the length and mass of the elemental string, respectively. By definition, linear mass density, μ is

The elemental string AB has a curvature which looks like an arc of a circle with centre at O, radius R and the arc subtending an angle θ at the origin O. The angle θ can be written in terms of arc length and radius as θ = \(\frac{d l}{R}\). The centripetal acceleration supplied by the tension in the string is

Then, centripetal force can be obtained when mass of the string (dm) is included in equation (3)

The centripetal force experienced by elemental string can be calculated by substituting equation (2) in equation (4) we get

The tension T acts along the tangent of the elemental segment of the string at A and B. Since the arc length is very small, variation in the tension force can be ignored. We can resolve T into horizontal component T cos \(\left(\frac{\theta}{2}\right)\) and vertical component T sin \(\left(\frac{\theta}{2}\right)\) The horizontal component at A and B are equal in magnitude but opposite in direction; therefore, they cancel each other. Since the elemental arc length AB is taken to be very small, the vertical components at A and B appears to acts Vertical towards the centre of the arc and hence, they add up. The net radial force F_r is

Since the amplitude of the wave is very small when it is compared with the length of the spring, the sine of small angle is approximated as sin \(\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}\). Hence equation (6) can be written as

Applying Newton’s second law to the elemental string in the radial direction, under equilibrium, the radial component of the force is equal to the centripetal force. Hence equating equation (5)

Question 4.
Describe Newton’s formula for velocity of sound waves in air and also discuss the Laplace’s correction.
Answer:
Newton’s formula for speed of sound waves in air: Sir Isaac Newton assumed that when sound propagates in air, the formation of compression and rarefaction takes place in a very slow manner so that the process is isothermal in nature. That is, the heat produced during compression (pressure increases, volume decreases), and heat lost during rarefaction (pressure decreases, volume increases) occur over a period of time such that the temperature of the medium remains constant. Therefore, by treating the air molecules to form an ideal gas, the changes in pressure and volume obey Boyle’s law, Mathematically
PV = constant …(1)
Differentiating equation (1), we get

where, B_T is an isothermal bulk modulus of air. Substituting equation (2) in equation the speed of sound in air is

Since P is the pressure of air whose value at NTP (Normal Temperature and Pressure) is 76 cm of mercury, we have

Here ρ is density of air, then the speed of sound in air at Normal Temperature and Pressure (NTP) is

But the speed of sound in air at 0°C is experimentally observed as 332 m s^-1 which is close upto 16% more than theoretical value (Percentage error is ). This error is not small.
Laplace’s correction: In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is

Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47.

Question 5.
Write short notes on reflection of sound waves from plane and curved surfaces. Reflection of sound through the plane surface
Answer:
When the sound waves hit the plane wall, they bounce off in a manner similar to that of light. Suppose a loudspeaker is kept at an angle with respect to a wall (plane surface), then the waves coming from the source (assumed to be a point source) can be treated as spherical wave fronts (say, compressions moving like a spherical wave front). Therefore, the reflected wave front on the plane surface is also spherical, such that its centre of curvature (which lies on the other side of plane surface) can be treated as the image of the sound source (virtual or imaginary loud speaker) which can be assumed to be at a position behind the plane surface.

Reflection of sound through the curved surface: The behaviour of sound is different when – it is reflected from different surfaces-convex or concave or plane. The sound reflected from a convex surface is spread out and so it is easily attenuated and weakened. Whereas, if it is reflected from the concave surface it will converge at a point and this can be easily amplified. The parabolic reflector (curved reflector) which is used to focus the sound precisely to a point is used in designing the parabolic mics which are known as high directional microphones.

We know that any surface (smooth or rough) can absorb sound. For example, the sound produced in a big hall or auditorium or theatre is absorbed by the walls, ceilings, floor, seats etc. To avoid such losses, a curved sound board (concave board) is kept in front of the speaker, so that the board reflects the sound waves of the speaker towards the audience. This method will minimize the spreading of sound waves in all possible direction in that hall and also enhances the uniform distribution of sound throughout the hall. That is why a person sitting at any position in that hall can hear the sound without any disturbance.

Question 6.
Briefly explain the concept of superposition principle.
Answer:
Superposition Principle: When a jerk is given to a stretched string which is tied at one end, a wave pulse is produced and the pulse travels along the string. Suppose two persons holding the stretched string on either side give a jerk simultaneously, then these two wave pulses move towards each other, meet at some point and move away from each other with their original identity. Their behaviour is very different only at the crossing/meeting points; this behaviour depends on whether the two pulses have the same or different shape.
When the pulses have the same shape, at the crossing, the total displacement is the algebraic sum of their individual displacements and hence its net amplitude is higher than the amplitudes of the individual pulses. Whereas, if the two pulses have same amplitude but shapes are 180° out of phase at the crossing point, the net amplitude vanishes at that point and the pulses will recover their identities after crossing.

Only waves can possess such a peculiar property and it is called superposition of waves. This means that the principle of superposition explains the net behaviour of the waves when they overlap. Generalizing to any number of waves i.e., if two are more waves in a medium move simultaneously, when they overlap, their total displacement is the vector sum of the individual displacements. We know that the waves satisfy the wave equation which is a linear second order homogeneous partial differential equation in both space coordinates and time. Hence, their linear combination (often called as linear superposition of waves) will also satisfy the same differential equation. To understand mathematically, let us consider two functions which characterize the displacement of the waves, for example,
y₁ = A₁ sin (kx – ωt) and y₂= A₂ cos (kx – ωt)
Since, both y₁ and y₂ satisfy the wave equation (solutions of wave equation) then their algebraic sum
y = y₁ + y₂
also satisfies the wave equation. This means, the displacements are additive. Suppose we multiply y₁ and y₂ with some constant then their amplitude is scaled by that constant Further, if C₁ and C₂ are used to multiply the displacements y₁ and y₂, respectively, then, their net displacement y is
C = C₁y₁ + C₂y₂
This can be generalized to any number of waves. In the case of n such waves in more than one dimension the displacements are written using vector notation. Here, the net displacement \(\vec{y}\) is

The principle of superposition can explain the following :
(a) Space (or spatial) Interference (also known as Interference)
(b) Time (or Temporal) Interference (also known as Beats)
(c) Concept of stationary waves
Waves that obey principle of superposition are called linear waves (amplitude is much smaller than their wavelengths). In general, if the amplitude of the wave is not small then they are called non-linear waves. These violate the linear superposition principle, e.g., laser. In this chapter, we will focus our attention only on linear waves.

Question 7.
Explain how the interference of waves is formed.
Answer:
Consider two harmonic waves having identical frequencies, constant phase difference cp and same wave form (can be treated as coherent source), hut having amplitudes A₁ and A₂, then

Suppose they move simultaneously in a particular direction, then interference occurs (i.e., overlap of these two waves). Mathematically
y = y₁ + y₂ …. (3)
Therefore, substituting equation (1) and equation (3) in equation (3), we get

By squaring and adding equation (5) and equation (6), we get

Since, intensity is square of the amplitude (I = A₂), we have

This means the resultant intensity at any point depends on the phase difference at that point.

(a) For constructive interference:
When crests of one wave overlap with crests of another wave, their amplitudes will add up and we get constructive interference. The resultant wave has a larger amplitude than the individual waves as shown in figure (a).

The constructive interference at a point occurs if there is maximum intensity at that point, which means that

This is the phase difference in which two waves overlap to give constructive interference. Therefore, for this resultant wave,

(b) For destructive interference: When the trough of one wave overlaps with the crest of another wave, their amplitudes “cancel” each other and we get destructive interference as shown in figure (b). The resultant amplitude is nearly zero. The destructive interference occurs if there is minimum intensity at that point, which means cos φ = – 1 ⇒ φ = π, 3π, 5π,… = (2n – 1) K, where n = 0, 1, 2, …. i.e. This is the phase difference in which two waves overlap to give destructive interference. Therefore,

Hence, the resultant amplitude A = |A₁ – A₂|

Question 8.
Describe the formation of beats.
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second
n = |f₁ – f₂|

Question 9.
What are stationary waves?. Explain the formation of stationary waves and also write down the characteristics of stationary waves.
Answer:
Explanation of stationary waves: When the wave hits the rigid boundary it bounces back to the original medium and can interfere with the original waves. A pattern is formed, which are known as standing waves or stationary waves.
Explanation: Consider two harmonic progressive waves (formed by strings) that have the same amplitude and same velocity but move in opposite directions. Then the displacement of the first wave (incident wave) is
y₁ = A sin (kx – ωt) (waves move toward right) …(1)
and the displacement of the second wave (reflected wave) is
y₂ = A sin (kx + ωt) (waves move toward left) …(2)
both will interfere with each other by the principle of superposition, the net displacement is
y = y₁ + y₂ …… (3)
Substituting equation (1) and equation (2) in equation (3), we get
y = A sin (kx – ωt) + A sin (kx + ωt) …(4)
Using trigonometric identity, we rewrite equation (4) as
y(x, t) = 2A cos (ωt) sin (kx) …(5)
This represents a stationary wave or standing wave, which means that this wave does not move either forward or backward, whereas progressive or travelling waves will move forward or backward. Further, the displacement of the particle in equation (5) can be written in more compact form,
y(x, t) = A’ cos (ωt)
where, A’ = 2A sin (foe), implying that the particular element of the string executes simple harmonic motion with amplitude equals to A’. The maximum of this amplitude occurs at positions for which

where m takes half integer or half integral values. The position of maximum amplitude is known as antinode. Expressing wave number in terms of wavelength, we can represent the anti-nodal positions as

The distance between two successive antinodes can be computed by

Similarly, the minimum of the amplitude A’ also occurs at some points in the space, and these points can be determined by setting
sin (kx) = 0 ⇒ kx = 0, π, 2π, 3π, … = nπ
where n takes integer or integral values. Note that the elements at these points do not vibrate (not move), and the points are called nodes. The n^th nodal positions is given by,

Characteristics of stationary waves:

1. Stationary waves are characterised by the confinement of a wave disturbance between two rigid boundaries. This means, the wave does not move forward or backward in a medium (does not advance), it remains steady at its place. Therefore, they are called “stationary waves or standing waves”.
2. Certain points in the region in which the wave exists have maximum amplitude, called as anti-nodes and at certain points the amplitude is minimum or zero, called as nodes.
3. The distance between two consecutive nodes (or) anti-nodes is \(\frac{\lambda}{2}\)
4. The distance between a node and its neighbouring anti-node is \(\frac{\lambda}{4}\)
5. The transfer of energy along the standing wave is zero.

Question 10.
Discuss the law of transverse vibrations in stretched strings.
Answer:
Laws of transverse vibrations in stretched strings: There are three laws of transverse vibrations of stretched strings which are given as follows:
(i) The law of length: For a given wire with tension T (which is fixed) and mass per unit length µ (fixed) the frequency varies inversely with the vibrating length. Therefore,

⇒ l × f = C, where C is a constant

(ii) The law of tension: For a given vibrating length l (fixed) and mass per unit length p , (fixed) the frequency varies directly with the square root of the tension T,

(iii) The law of mass: For a given vibrating length l (fixed) and tension T (fixed) the frequency varies inversely with the square root of the mass per unit length µ,

Question 11.
Explain the concepts of fundamental frequency, harmonics and overtones in detail. Fundamental frequency and overtones in detail.
Answer:
Fundamental frequency and overtones: Let us now keep the rigid boundaries at x = 0 and x = L and produce a standing waves by wiggling the string (as in plucking strings in a guitar). Standing waves with a specific wavelength are produced. Since, the amplitude must vanish at the boundaries, therefore, the displacement at the boundary must satisfy the following conditions
x(x = 0, t) = 0 and y(x = L, t) = 0
Since, the nodes formed at a distance \(\frac{\lambda_{n}}{2}\) apart, we have \(n\left(\frac{\lambda_{n}}{2}\right)=\mathrm{L}\), where n is an integer, L is the length between the two boundaries and λ_n is the specific wavelength that satisfy the specified boundary conditions. Hence,

Therefore, not all wavelengths are allowed. The (allowed) wavelengths should fit with the specified boundary conditions, i.e., for n = 1, the first mode of vibration has specific wavelength λ₁ = 2L. Similarly for n = 2, the second mode of vibration has specific wavelength

For n = 3, the third mode of vibration has specific wavelength

The frequency of each mode of vibration (called natural frequency) can be calculated.

The lowest natural frequency is called the fundamental frequency.

The second natural frequency is called the first over tone.

The third natural frequency is called the second over tone.

Therefore, the n^th natural frequency can be computed as integral (or integer ) multiple of fundamental frequency, i.e.,
f_n = nf₁ where n is an integer …(5)
If natural frequencies are written as integral multiple of fundamental frequencies, then the frequencies are called harmonics. Thus, the first harmonic is f₁ = f₁ (the fundamental frequency is called first harmonic), the second harmonic is f₂ = 2f₁, the third harmonic is f₃ = 3f₁ etc.

Question 12.
What is a sonometer? Give its construction and working. Explain how to determine the frequency of tuning fork using sonometer.
Answer:
Stationary waves in sonometer: Sono means sound related, and sonometer implies sound-related measurements.
It is a device for demonstrating the relationship between the frequency of the sound produced in the transverse standing wave in a string, and the tension, length and mass per unit length of the string. Therefore, using this device, we can determine the following quantities:

(a) the frequency of the tuning fork or frequency of alternating current
(b) the tension in the string
(c) the unknown hanging mass
Construction: The sonometer is made up of a hollow box which is one meter long with a uniform metallic thin string attached to it. One end of the string is connected to a hook and the other end is connected to a weight hanger through a pulley as shown in figure. Since only one string is used, it is also known as monochord. The weights are added to the free end of the wire to increase the tension of the wire. Two adjustable wooden knives are put over the board, and their positions are adjusted to change the vibrating length of the stretched wire.
Working: A transverse stationary or standing wave is produced and hence, at the knife edges P and Q, nodes are formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is then

Let f be the frequency of the vibrating element, T the tension of in the string and p the mass per unit length of the string. Then using equation, we get

Let ρ be the density of the material of the string and d be the diameter of the string. Then the mass per unit length μ

Question 13.
Write short notes on intensity and loudness.
Answer:
Intensity and loudness: Consider a source and two observers (listeners). The source emits sound waves which carry energy. The sound energy emitted by the source is same regardless of whoever measures it, i.e., it is independent of any observers standing in that region. But the sound received by the two observers may be different; this is due to some factors like sensitivity of ears, etc. To quantify such thing, we define two different quantities known as intensity and loudness of sound.

Intensity of sound: When a sound wave is emitted by a source, the energy is carried to all possible surrounding points. The average sound energy emitted or transmitted per unit time or per second is called sound power.
Therefore, the intensity of sound is defined as “the sound power transmitted per unit area taken normal to the propagation of the sound wave ”.
For a particular source (fixed source), the sound intensity is inversely proportional to the square of the distance from the source.

This is known as inverse square law of sound intensity.
the power output does not depend on the observer and depends on the baby. Therefore, Loudness of sound: Two sounds with same intensities need not have the same loudness. For example, the sound heard during the explosion of balloons in a silent closed room is very loud when compared to the same explosion happening in a noisy market. Though the intensity of the sound is the same, the loudness is not. If the intensity of sound is increased then loudness also increases. But additionally, not only does intensity matter, the internal and subjective experience of “how loud a sound is” i.e., the sensitivity of the listener also matters here. This is often called loudness. That is, loudness depends on both intensity of sound wave and sensitivity of the ear (It is purely observer dependent quantity which varies from person to person) whereas the intensity of sound does not depend on the observer. The loudness of sound is defined as “the degree of sensation of sound produced in the ear or the perception of sound by the listener”.

Question 14.
Explain how overtones are produced in a:
(a) Closed organ pipe
(b) Open organ pipe
Answer:
(a) Closed organ pipes: Clarinet is an example of a closed organ pipe. It is a pipe with one end closed and the other end open. If one end of a pipe is closed, the wave reflected at this closed end is 180° out of phase with the incoming wave. Thus there is no displacement of the particles at the closed end. Therefore, nodes are formed at the closed end and anti-nodes are formed at open end.

Let us consider the simplest mode of vibration of the air column called the fundamental mode. Anti-node is formed at the
open end and node at closed end. From the figure, let L be the length of the tube and the wavelength of the wave produced. For the fundamental mode of vibration, we have,

which is called the fundamental note

The frequencies higher than fundamental frequency can be produced by blowing air strongly at open end. Such frequencies are called overtones.

The figure (b) shows the second mode of vibration having two nodes and two antinodes,

is called first over tone, since here, the frequency is three times the fundamental frequency it is called third harmonic.
The figure (c) shows third mode of vibration having three nodes and three anti-nodes.

is called second over tone, and since n = 5 here, this is called fifth harmonic. Hence, the closed organ pipe has only odd harmonics and frequency of the n^th harmonic is f_n = (2n + 1)f₁. Therefore, the frequencies of harmonics are in the ratio
f₁ : f₂ : f₃ : f₄ …… = 1 : 3 : 5 : 7 : …… ……… (3)

(b) Open organ pipes: Flute is an example of open organ pipe. It is a pipe with both the ends open. At both open ends, anti-nodes are formed. Let us consider the simplest mode of vibration of the air column called fundamental mode. Since anti-nodes are formed at the open end, a node is formed at the mid-point of the pipe.

From figure (d), if L be the length of the tube, the wavelength of the wave produced is given by

The frequency of the note emitted is

which is called the fundamental note. The frequencies higher than fundamental frequency can be produced by blowing air strongly at one of the open ends. Such frequencies are called overtones.

The Figure (e) shows the second mode of vibration in open pipes. It has two nodes and three anti-nodes, and therefore,

is called first over tone. Since n = 2 here, it is called second harmonic.
The Figure (f) above shows the third mode of vibration having three nodes and four anti-nodes.


is called second over tone. Since n = 3 here, it is called the third harmonic.
Hence, the open organ pipe has all the – harmonics and frequency of nth harmonic is f_n = nf₁. Therefore, the frequencies of harmonics are in the ratio
f₁ : f₂ : f₃ : f₄ … = 1 : 2 : 3 : 4 : … …(6)

Question 15.
How will you determine the velocity of sound using resonance air column apparatus?
Answer:
Resonance air column apparatus:
The resonance air column apparatus and first, second and third resonance The resonance air column apparatus is one of the simplest techniques to measure the speed of sound in air at room temperature. It consists of a cylindrical glass tube of one meter length whose one end A is open and another end B is connected to the water reservoir R through a rubber tube as shown in figure. This cylindrical glass tube is mounted on a vertical stand with a scale attached to it. The tube is partially filled with water and the water level can be adjusted by raising or lowering the water in the reservoir R. The surface of the water will act as a closed erid and other as the open end. Therefore, it behaves like a closed organ pipe, forming nodes at the surface of water and antinodes at the closed end. When a vibrating tuning fork is brought near the open end of the tube, longitudinal waves are formed inside the air column. These waves move downward as shown in Figure, and reach the surfaces of water and get reflected and produce standing waves. The length of the air column is varied by changing the water level until a loud sound is produced in the air column. At this particular length the frequency of Waves in the air column resonates with the frequency of the tuning fork (natural frequency of the tuning fork). At resonance, the frequency of sound waves produced is equal to the frequency of the tuning fork. This will occur only when the length of air column is proportional to \(\left(\frac{1}{4}\right)^{t h}\) of the wavelength of the sound waves produced. Let the first resonance occur at length L₁, then

But since the antinodes are not exactly formed at the open end, we have to include a correction, called end correction e, by assuming that the antinode is formed at some small distance above the open end. Including this end correction, the first resonance is

Now the length of the air column is increased to get the second resonance. Let L₂ be the length at which the second resonance occurs. Again taking end correction into account, we have

In order to avoid end correction, let us take the difference of equation (3) and equation (2)

The speed of the sound in air at room temperature can be computed by using the formula

Further, to compute the end correction, we use equation (2) and equation (3), we get

Question 16.
What is meant by Doppler effect? Discuss the following cases
(1) Source in motion and Observer at rest
(a) Source moves towards observer
(b) Source moves away from the observer
(2) Observer in motion and Source at rest.
(a) Observer moves towards Source
(b) Observer resides away from the Source
(3) Both are in motion
(a) Source and Observer approach each other
(b) Source and Observer resides from each other
(c) Source chases Observer
(d) Observer chases Source
Answer:
Doppler Effect: When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.
1. Source in motion and the observer at rest
(a) Source moves towards the observer: Suppose a source S moves to the right (as shown in figure) with a velocity v_s and let the frequency of the sound waves produced by the source be fs. We assume the velocity of sound in a medium is v.

The compression (sound wave front) produced by the source S at three successive instants of time are shown in the figure. When S is at position x₁ the compression is at C₁. When S is at position x₂, the compression is at C₂ and similarly for x₃ and C₃. Assume that if reaches the observer’s position A then at that instant C₂ reaches the point B and C₃ reaches the point C as shown in the figure. It is obvious to see that the distance between compressions C₂ and C₃ is shorter than distance between C₁ and C₂. This means the wavelength decreases when the source S moves towards the observer O (since sound travels longitudinally and wavelength is the distance between two consecutive compressions). But frequency is inversely related to wavelength and therefore, frequency increases.
Let λ be the wavelength of the source S as measured by the observer when S is at position x₁ and λ’ be wavelength of the source observed by the observer when S moves to position x₂. Then the change in wavelength is ∆λ = λ – λ’ = v_s t, where t is the time taken by the source to travel between x₁ and x₂. Therefore,

On substituting equation (2) in equation (1), we get

(b) Source moves away from the observer: Since the velocity here of the source is opposite in direction when compared to case (a), therefore, changing the sign of the velocity of the source in the above case i.e, by substituting (v_s ➝ – v_s) in equation (1), we get

Using binomial expansion again, we get,

2. Observer in motion and source at rest:
(a) Observer moves towards Source:
Let us assume that the observer O moves towards the source S with velocity v_o. The source S is at rest and the velocity of sound waves (with respect to the medium) produced by the source is v. From the figure, we observe that both vo and v are in opposite direction. Then, their relative velocity is v_r = v + v_o. The wavelength of the sound wave is \(\lambda=\frac{v}{f}\), which means the frequency observed by the observer O is f’ = \(\). Then

(b) Observer recedes away from the Source: If the observer O is moving away (receding away) from the source S, then velocity v₀ and v moves in the same direction. Therefore, their relative velocity is v_r = v – v_r. Hence, the frequency observed by the observer O is

3. Both are in motion:
(a) Source and observer approach each other:

Let v_s and v_o be the respective velocities of source and observer approaching each other as shown in figure. In order to calculate the apparent frequency observed by the observer, as a simple calculation, let us have a dummy (behaving as observer or source) in between the source and observer. Since the dummy is at rest, the dummy (observer) observes the apparent frequency due to approaching source as given in equation (3) as

At that instant of time, the true observer approaches the dummy from the other side. Since the source (true source) comes in a direction opposite to true observer, the dummy (source) is treated as stationary source for the true observer at that instant. Hence, apparent frequency when the true observer approaches the stationary source (dummy source), from equation (7) is

Since this is true for any arbitrary time, therefore, comparing equation (9) and equation (10), we get

(b) Source and observer resides from each other

Here, we can derive the result as in the previous case. Instead of a detailed calculation, by inspection from figure, we notice that the velocity of the source and the observer each point in opposite directions with respect to the case in (a) and hence, we substitute (v_s ➝ – v_s) and (v₀ ➝ – v_o) in equation (11), and therefore, the apparent frequency observed by the observer when the source and observer recede from each other is

(c) Source chases the observer

Only the observer’s velocity is oppositely directed when compared to case (a).

(d) Observer chases the source

Only the source velocity is oppositely directed when compared to case (a). Therefore, substituting v_s ➝ – v_s in equation (12), we get

Samacheer Kalvi 11th Physics Waves Numerical Problems

Question 1.
The speed of a wave in a certain medium is 900 m/s. If 3000 waves passes over a certain point of the medium in 2 minutes, then compute its wavelength?
Answer:
Speed of the wave in medium v = 900 ms^-1

Question 2.
Consider a mixture of 2 mol of helium and 4 mol of oxygen. Compute the speed of sound in this gas mixture at 300 K.
Answer:
The mixture of helium and oxygen.

Question 3.
A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bed rock and returns to the ship after 3.5 s. After the ship moves to 100 km it sends another signal which returns back after 2s. Calculate the depth of the sea in each case and also compute the difference in height between two cases.
Answer:
Speed of SONAR waves in water c = 1500 ms^-1
Time taken to reflect from the bottom of the sea, 2t = 3.5 sec
∴ t = 1.75 sec
Distance covered in forward and reflected backward (d₁) = c × t
d₂ = 1500 × 1.75 = 2625 m
After ship moves in a distance = 150 km
Time taken to reflect by the waves 2t = 2s
t = 1s
Distance covered by the waves (d₂) = c × t = 1500 × 1 = 1500 m
The different between the height of two cases = 2625 – 1500
h_difference = 1124 m

Question 4.
A sound wave is transmitted into a tube as shown in figure. The sound wave splits into two waves at the point A which recombine at point B. Let R be the radius of the semicircle which is varied until the first minimum. Calculate the radius of the semi-circle if the wavelength of the sound is 50.0 m
Answer:
The sound travelling in the curved path distance = πR
L₁ = πR
The sound travelling in the straight path distance = 2R
L₂ = 2R

The path distance of straight and curved path AP = L₁ – L₂

Question 5.
N tuning forks are arranged in order of increasing frequency and any two successive tuning forks give n beats per second when sounded together. If the last fork gives double the frequency of the first (called as octave), Show that the frequency of the first tuning fork is f = (N – 1)n.
Answer:
Total number of fork = N
The frequency of the 1st fork = f
The frequency of the last fork = 2f
∴ a_n = a + (n – 1)d
2f = f + (N – 1)n
2f – f = (N – 1)n
∴ f = (N – 1)n

Question 6.
Let the source propagate a sound wave whose intensity at a point (initially) be I. Suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. Calculate now the new intensity of sound at the same point?
Answer:
Intensity of sound wave (old) = I₁
Amplitude of sound wave (A₂) = 2A₁
Frequency of the sound wave I₂ = ?

Question 7.
Consider two organ pipes of same length in which one organ pipe is closed and another organ pipe is open. If the fundamental frequency of closed pipe is 250 Hz. Calculate the fundamental frequency of the open pipe.
Answer:
Fundamental frequency of closed organ pipe

Fundamental frequency:f open organ pipe \(f_{o}=\frac{v}{2 l}=?\)

Question 8.
A police in a siren car moving with a velocity 20 ms^– chases a thief who is moving in a car with a velocity v₀ ms^-1. The police car sounds at frequency 300 Hz, and both of them move towards a stationary siren of frequency 400 Hz. Calculate the speed in which thief is moving.
(Assume the thief does not observe any beat)
Answer:
Velocity of sound v = 330 ms^-1
Velocity of car (v_s ) = 20 ms^-1
Frequency of car (f₁) = 300 Hz
Frequency of stationary siren (f₂) = 400 Hz
The speed of the thief (v_o ) = ?

330 – v_o = 413.325 + 1.2525v_o
2.2525v_o = – 83. 325
v_o = – 36.99
∴ speed of the thief in moving = 36.99 ms^-1

Question 9.
Consider the following function:
(a) y = x² + 2 α tx
(b) y = (x + vt)² which among the above function can be characterized as a wave?
Answer:
(a) y = x² + 2 α tx. This expression is not a wave equation.
(b) y = (x + vt)² . This expression is satisfies the wave equation.

Samacheer Kalvi 11th Physics Waves Conceptual Questions

Question 1.
Why is it that transverse waves cannot be produced in a gas? Can the transverse waves can be produced in solids and liquids?
Answer:
Transverse waves travel in the form of crests and through. They involve changes in the shape of the medium. As gas has no elasticity of shape, hence transverse waves cannot be produced in it. So, they can be transmitted through media which sustain shearing stress such as solids, strings and liquid surface.

Question 2.
Why is the roar of our national animal different from the sound of a mosquito?
Answer:
Both sounds travel at the speed of sound. The speed of sound varies according to the density and temperature of the air, but not according to the loudness of the sound,at least not for the levels of loudness we talking about here.
The roaring of the lion will be audible a lot further away, but that’s simply because it’s louder.

Question 3.
A sound source and listener are both stationary and a strong wind is blowing. Is there a Doppler effect?
Answer:
Yes, It does not matter whether the sound source or the transmission media are in motion, vibrations will be compressed in the direction of convergence and dilated in the direction of divergence.

Question 4.
In an empty room why is it that a tone sounds louder than in the room having things like furniture etc.
Answer:
Because in a furniture room will absorb the sound waves, hence there went be any echo. But in an empty room reflect the sound. Therefore there will be echo hence we hear sound louder.

Question 5.
How do animals sense impending danger of hurricane?
Answer:
Some animals are believed to be sensitive to be low frequency sound waves emitted by hurricanes, they can also detect the slight drops in air and water pressure that signal a storm’s approach.

Question 6.
Is it possible to realize whether a vessel kept under the tap is about to fill with water?
Answer:
The frequency of the note produced by an air column is inversely proportional to its length. As the level of water is the vessel rises, the length of the air column above it decreases. It produces sound of decreasing frequency, i.e., the sound becames shorter. From the shrillness of sound, it is possible to realize whether the vessel is filled which water.
v_min = 11.71 ms^-1

Samacheer Kalvi 11th Physics Waves Textual Evaluation Solved Additional Questions Solved

I. Choose the correct answer from the following:

Question 1.
Mechanical Waves
(a) are longitudinal only
(b) are transverse only
(c) can be both longitudinal and transverse.
(d) are neither longitudinal for transverse waves.
Answer:
(c) can be both longitudinal and transverse.

Question 2.
Sound whose frequency is 50 Hz?
(a) has a relatively short wavelength.
(b) has a relatively long wavelength
(c) is very loud
(d) is very intense
Answer:
(a) has a relatively short wavelength.

Question 3.
Sound travels fastest in …….
(a) Steel
(b) air
(c) water
(d) vaccum
Answer:
(a) steel

Question 4.
A boat at anchor is rocked by waves of velocity 25m/s, having crests 100 m apart. The boat bounches up once in every
(a) 4.0s
(b) 2500s
(c) 0.25s
(d) 75s
Answer:
(a) 4.0s
Hint:
λ = distance between crests = 100 m frequency v = \(\frac{25}{100}=\frac{1}{4} \mathrm{s}^{-1}\)
Therefore, the crests reach the boat once every 4 seconds.

Question 5.
Choose the correct statement:
(a) sound waves are transverse waves
(b) sound travels fastest through vaccum.
(c) sound travels faster in solids than in gases.
(d) sound travels faster in gases than in liquids.
Answer:
(c) sound travels faster in solids than in gases.

Question 6.
Transverse waves can propagate ……
(a) both in a gas and in a metal
(b) in a gas but not in a metal
(c) not in a gas but in a metal
(d) neither in a gas nor in a metal
Answer:
(a) not in a gas but in a metal.

Question 7.
The speed of the wave represented by y = A sin(ωt – kx) is ……..
(a) k/ω
(b) ω/k
(c) ωk
(d) 1/ωk
Answer:
(b) ω/k

Question 8.
The equation of a wave travelling in a string can be written as y = 3 cos {π(100t – x)} where y and x are in cm and t is in seconds. Then the value of wavelength is …….
(a) 100 cm
(b) 2 cm
(c) 50 cm
(d) 4 cm
Answer:
(b) 2 cm
Hint:
On comparing given equation with y = A cos (kx – ωt), we get

Question 9.
A wave of frequency 500 Hzhas a velocity 300 m/s. The distance between two nearest points which are 60° out of phase, is ……
(a) 0.2 m
(b) 0.1 m
(c) 0.4 m
(d) 0.5 m
Answer:
(a) 0.1 cm

Question 10.
The equation of a wave travelling on a string is , where x, y are in cm and t in seconds. The velocity of the waves is ……..
(a) 64 cm/s in – x direction
(b) 32 cm/s in – x direction
(c) 32 cm/s in +x direction
(d) 64 cm/s in + x direction
Answer:
(d) 64 cm/s in + x direction to S

Question 11.
The equation of a wave is where y, x are in cm and t in seconds. The amplitude wavelength, velocity and frequency of the wave are, respectively, …….
(a) 4 cm, 32 cm, 16 cm/s, 0.5 Hz
(b) 8 cm, 16 cm, 32 cm/s, 1.0 Hz
(c) 4 cm, 32 cm, 32 cm/s, 0.5 Hz
(d) 8 cm, 16 cm, 16 cm/s, 1.0 Hz
Answer:
(a) 4 cm, 32 cm, 16 cm/s, 0.5 Hz

Question 12.
The diagram shows the profile of a wave, which of the following pairs of points are in phase?
(a) A, B
(b) B, C
(c) B, D
(d) B, E

Answer:
(d) B, E

Question 13.
Ultrasonic waves are those waves which ………
(a) human beings cannot hear
(b) human beings can hear
(c) have high velocity
(d) have large amplitude
Answer:
(a) human beings cannot hear

Question 14.
A transverse wave of amplitude 0.5m, wavelength 1 m and frequency 2Hz is propogating in a string in the negative x direction. The equation of this wave is ….
(a) y = 0.5 sin (2πx – 4πt)
(b) y = 0.5 sin (2πx + 4πt)
(c) y = 0.5 sin (πx – 2πt)
(d) y = 0.5 cos (kx – 2πt)
Answer:
(b) y = 0.5 sin (2πx + 4πt)
Hint:
y = A sin(kx + ωt)
Here A = 0.5 m

Question 15.
With the rise of temperature, the speed of sound in a gas ……..
(a) increases
(b) decreases
(c) remain the same
(d) may increase or decrease depending on the corresponding change in pressure.
Answer:
(a) increases

Question 16.
Speed of sound in a gas in proportional to …….
(a) square root of isothermal elasticity
(b) square root of adiabatic elasticity
(c) isothermal elasticity
(d) adiabatic elasticity
Answer:
(b) square root of adiabatic elasticity

Question 17.
The velocity of sound in are is not affected by change in the …….
(a) atmospheric pressure
(b) moisture content of air
(c) temperature of air
(d) composition of air
Answer:
(a) atmospheric pressure

Question 18.
A longitudinal wave is described by the equation y = y₀ sin 2π (ft – x/λ). The maximum particle velocity is equal to four times the wave velocity if ……..

Answer:

Question 19.
If v₀ and v denote the sound velocity and the rms velocity of the molecules in a gas, then ……..

Answer:

Question 20.
With the propagation of a longitudinal wave through a material medium, the quantities transferred in the direction of propagation are ……..
(a) energy, momentum and mass
(b) energy and momentum
(c) energy and mass
(d) energy
Answer:
(b) energy and momentum

Question 21.
If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity will ………
(a) increase by a factor of 2
(b) decrease by a factor of 2
(c) decrease by a factor of 4
(d) remain unchanged
Answer:
(c) decrease by a factor of 4

Question 22.
When a source of sound is in motion towards a stationary observer, the effect observed is
(a) increase in the velocity of sound only
(b) decrease in the velocity of sound only
(c) increase in frequency of sound only
(d) increase in both the velocity and the frequency of sound
Answer:
(c) increase in frequency of sound only

Question 23.
The apparent wavelength of the light from a star, moving away from the earth, is 0.01% more than its real wave length. The speed of the star with respect to the earth is ……
(a) 10 km/s
(b) 15 km/s
(c) 30 km/s
(d) 60 km/s
Answer:
(c) 30 km/s

Question 24.
The frequency of a radar is 780 MHz. When it is reflected from an approaching aeroplane the opponent frequency is more than the actual frequency by 2.6 kHz. The speed of the aeroplane is ……
(a) 0.25 km/s
(b) 0.5 km/s
(c) 1.0 km/s
(d) 2.0 km/s
Answer:
(b) 0.5 km/s

Question 25.
The temperature at which the speed of sound in air becomes double its value at 27°C is …….
(a) 54°C
(b) 327°C
(c) 927°C
(d) -123°C
Answer:
(c) 927°C

Question 26.
The equation of a transverse wave is given by y = 10 sin{π(0.01x – 2t)} where y and x are in cm and t is in seconds. Its frequency is …….
(a) 10 s^-1
(b) 2 s^-1
(c) 1 s^-1
(d) 0.01 s^-1
Answer:
(c) 1 s^-1
Hint:
Comparing with the standard equation y = A sin(kx – ωt),

Question 27.
When sound waves travel from air to water, which of the following remains constant?
(a) velocity
(b) frequency
(c) wavelength
(d) all of these
Answer:
(b) frequency

Question 28.
The speed of sound in oxygen is 332 m/s at STP. The speed of sound in hydrogen at STP will be ……..
(a) 53/2 m/s
(b) 2546 m/s
(c) 1328 m/s
(d) 664 m/s
Answer:
(c) 1328 m/s

Question 29.
If v_a, v_h, and v_m are the speeds of sound in air, hydrogen and a metal at the same temperature, then ……..
(a) v_h > v_a > v_m
(b) v_m > v_h > v_a
(c) v_h > v_m > v_a
(d) v_a > v_h > v_m
Answer:
(b) v_m > v_h > v_a

Question 30.
Ultrasonic waves can be detected by ……
(a) telephone
(b) Hebb’s method
(c) Kundt’s tube
(d) Quincke’s tube
Answer:
(c) Kundt’s tube

Question 31.
The velocity of sound in a gas depends on ….
(a) Wavelength only
(b) density and elasticity of gas
(c) intensity only
(d) amplitude and frequency
Answer:
(b) density and elasticity of gas

Question 32.
When sound waves travel from air to water, which of these remains constant?
(a) velocity
(b) wavelength
(c) frequency
(d) all the above
Answer:
(c) frequency

Question 33.
When a wave goes from one medium to another, there is a change in
(a) velocity
(b) amplitude
(c) wavelength
(d) all the above
Answer:
(d) all the above

Question 34.
The equation of a sound wave is y = 0.0015 sin (62.8x + 316t). Find the wave length of the above ……
(a) 0.2 units
(b) 0.3 units
(c) 0.1 units
(d) 0.15 units
Answer:
(c) 0.1 units

Question 35.
Red shift is an illustration of ……….
(a) low temperature emission
(b) high frequency absorption
(c) Doppler effect
(d) Same unknown Phenomenon.
Answer:
(c) Doppler effect

Question 36.
The ratio of the velocity of sound in a monatomic gas to that in a triatomic gas having same molar mass, under similar conditions of temperature and pressure, is ………
(a) 1.12
(6) 1.25
(c) 1.50
(d) 1.6
Answer:
(a) 1.12

Question 37.
Doppler shift in frequency does not depend upon …….
(a) the actual frequency of the wave
(b) the velocity of the source from the listener.
(c) the velocity of the source.
(d) the velocity of the observer.
Answer:
(b) the velocity of the source from the listener.

Question 38.
If the density of oxygen is 16 times that of hydrogen, what will be the ratio of the velocities of sound in them?
(a) 1 : 4
(b) 4 : 1
(c) 2 : 1
(d) 1 : 16
Answer:
(a) 1 : 4

Question 39.
Pitch of sound depends on ……
(a) frequency
(b) wavelength
(c) amplitude
(d) speed
Answer:
(a) frequency

Question 40.

Answer:

Question 41.
Which of the following equations represents a wave?
(a) y = A(ωt – kx)
(b) y = A sin ωt
(c) y = A cos kx
(d) y = A sin (at – bx + c)
Answer:
(d) y = A sin (at -bx + c)

Question 42.
A wave travels in a medium according to the equation of displacement given by y(x, t) = 0.03 sin{π(2t – 0.01 x)} where y and x are in metres and t in seconds. The wave length of the wave is …..
(a) 200 m
(b) 100 m
(c) 20 m
(d) 10 m
Answer:
(a) 200 m

Question 43.
The equation of a wave moving on string is y = 8 sin{π(0.002 x – 4t)} where x, y are in centimeter and t in seconds. The velocity of the wave is ……
(a) 100 cm/s
(b) 0.2π cm/s
(c) 4π cm/s
(d) 200 cm/s
Answer:
(d) 200 cm/s

Question 44.
If the velocity of sound in air is 340 ms^-1, a person singing a note of frequency 250 cps is producing sound waves with a wavelength of ……..
(a) 0.7
(b) 1.36 cm
(c) 1.36 m
(d) 85 km
Answer:
(c) 1.36

Question 45.
Asa transverse wave strikes against a fixed end …….
(a) its phase changes by 180°, but velocity does not change.
(b) its phase does not change, but velocity changes
(c) its velocity changes and phase too changes by 180°
(d) nothing can be predicted about changes in its velocity and phase.
Answer:
(a) its phase changes by 180°, but velocity does not change

Question 46.
A source of sounds is travelling with a velocity of 40 km/hr towards an observer and emits sound of frequency 2000 Hz. If the velocity of sound is 1220 km/hr, then what is the apparent frequency heard by the observer?
(a) 2068 Hz
(b) 2180 Hz
(c) 2000 Hz
(d) 1980 Hz
Answer:
(a) 2068 Hz

Question 47.
A vehicle with a horn of frequency n is moving with a velocity of 30m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n₁. Then (if the sound velocity in air is 300 m/s)
(a) n₁ = 10 n
(b) n₁ = 0
(c) n₁ = – 0.1 n
(d) n₁ = 0.1 n
Answer:
(b) n₁ = 0
Hint:
No Doppler effect is observed if the source moves perpendicular to the line joining the source and the observer. Therefore, the correct choice is (b).

Question 48.
The Doppler effect is applicable for ……..
(a) light waves
(b) sound waves
(c) space waves
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 49.
The speed of a wave in a medium is 760 m/s. If 3600 waves are passing through a point in the medium in 2 minutes, then its wavelength is ……
(a) 13.8 m
(b) 25.3 m
(c) 41.5 m
(d) 57.2 m
Answer:
(b) 25.3 m

Question 50.
If a sound wave travels from air to water, the quantity that remain unchanged is …….
(a) velocity
(b) wavelength
(c) frequency
(d) amplitude
Answer:
(c) frequency

Question 51.
Asa spherical wave propagates, …….
(a) the wave intensity remains constant
(b) the wave intensity decrease as the inverse of the distance from the source
(c) the wave intensity decreases as the inverse square of the distance from the source.
(d) The wave intensity decreases as the inverse cube of the distance from the source.
Answer:
(c) The wave intensity decreases as the inverse square of the distance from the source.

Question 52.
A source of sound and a listener are approaching each other with a speed of 40ms^-1.The apparent frequency of a note produced by the source is 400 Hz. Then its true frequency is (velocity of sound in air = 360 ms^-1)
(a) 320 Hz
(b) 400 Hz
(c) 360 Hz
(d) 420 Hz
Answer:
(a) 320 Hz

Question 53.
Sound waves of wavelength greater than that of audible sound are called ……
(a) infrasonic waves
(b) ultrasonic waves
(c) sonic waves
(d) seismic waves
Answer:
(a) infrasonic waves

Question 54.
The frequency of a sound wave is/and its velocity is v. If the frequency is increased to 4 f the velocity of the wave will be:
(a) v
(b) 2v
(c) 4 v
(d) v/4
Answer:
(a) v
Hint:
The velocity is a characteristic of the medium and, therefore, it remains constant.

Question 55.
Which of the following statement is untrue? The velocity of sound in a gas …….
(a) is independent of pressure
(b) increases with increase in temperature
(c) is dependent on molecular weight
(d) is greater in dry air than in moist air
Answer:
(d) is greater in dry air than in moist air

Question 56.
When a stone is dropped on the surface of still water, the waves produced are …….
(a) transverse
(b) longitudinal
(c) Stationary
(d) partly longitudinal and partly transverse.
Answer:
(d) Partly longitudinal and partly transverse.

Question 57.
The equation of a wave is y = 0.1 sin (100πt – kx) where x, y are in metres and t in seconds. If – the velocity of the wave is 100 m/s, then the value of k is
(a) 1 m^-1
(b) 2m^-1
(c) πm^-1
(d) 2πm^-1
Answer:
(c) πm^-1

Question 58.
A transverse wave propagating on a stretched string of linear density 3 × 10^-4 kg m^-1 is represented by the equation, y = 0.2 sin (1.5x + 60t)
Where x is in metres and t is in seconds. The tension in the string (in newtons) is:
(a) 0.24
(b) 0.48
(c) 1.20
(d) 1.80
Answer:
(a) 0.48

Question 59.
A transverse wave propagating along x-axis is represented by

where x is in metres and t is in seconds. The speed of the wave is ……….

Answer:
(c) 8m/s
Hint:

Question 60.
Two waves represented by the following equation are travelling in the same medium: y₁ = 5 sin 2π (75t – 0.25 x) and y₂ = 10 sin 2π (150 – 0.25x) The intensity ratio of the two waves is ……..
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
Answer:
(b) 1 : 4

Question 61.
A point source emits sound equally in all direction is a non-absorbing medium. Two points P and Q are at distances of 2m and 3m, respectively, from the source. The ratio of the intensities of the waves at P and Q is …….
(a) 3 : 2
(b) 4 : 9
(c) 2 : 3
(d) 9 : 4
Answer:
(d) 9 : 4

Question 62.
The waves produced by a motor boat sailing in water are ……..
(a) transverse
(b) longitudinal
(c) longitudinal and transverse
(d) stationary
Answer:
(c) longitudinal and transverse

Question 63.
Doppler effect in sound is due to ………
(a) motion of source
(b) motion of observer
(c) relative motion of source and observer
(d) none of the above
Answer:
(c) relative motion of source and observer

Question 64.
The velocity of sound in air at NTP is 330m/s. What will be its value when temperature is doubled and pressure is halved?
(a) 165 m/s
(b) 330 m/s
(c) 330 /\(\sqrt{2}\)
(d) 300/\(\sqrt{2}\) m/s
Answer:
(c) 330 /\(\sqrt{2}\)
Hint:
There is no effect of change of pressure on the velocity of sound in air. Further, v ∝ \(\sqrt{\mathrm{T}}\)

Question 65.
Sound waves travel at 350 m/s through warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air
(a) increases by a factor 20
(b) increases by a factor 10
(c) decreases by a factor 20
(d) decreases by a factor 10
Answer:
(b) increase by a factor 10
Hint:
Since the frequency remains the same, we have

Question 66.
A train moving at a speed of 220 m/s towards a stationary object, emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is
(a) 3000 Hz
(b) 3500 Hz
(c) 4000 Hz
(d) 5000 Hz
Answer:
(d) 5000 Hz

Question 67.
A source of sounds emitting waves of frequency 100 Hz and an observer O are located at same distance from each other The source is moving with a speed of 19.4 ms^-1 at an angle of 60° with the source-observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 ms^-1) is ……

(a) 97 Hz
(b) 100 Hz
(c) 103 Hz
(d) 106 Hz
Answer:
(c) 103 Hz

Question 68.
Beats occur because of
(a) interference
(b) reflection
(c) refraction
(d) Doppler effect
Answer:
(a) interference

Question 69.
A vibrating stretched string resonates with a tuning fork of frequency 512 Hz when the length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be ……
(a) 0.25 m
(b) 0.75 m
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m

Question 70.
A cylindrical tube, open at both ends has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now …….
(a) f/2
(b) f
(c) 3f/4
(d) 2f
Answer:
(a) f
Hint:
When the tube is dipped in water, it become a closed pipe of length L/2. Its fundamental frequency is

Question 71.
With the increase in temperature, the frequency of the found from an organ pipe
(a) decrease
(b) increase
(c) remains unchanged
(d) changes erractically
Answer:
(b) increase
Hint:
Frequency ∝ v/L. Now v and L both increase with temperature but increase of v is much more than the increase of L which is negligible. Thus frequency increases with temperature.

Question 72.
Two waves of the same frequency and amplitude super impose to produce a resultant disturbance of the same amplitude. The phase difference between the waves is ……
(a) zero
(b) π/3
(c) π/4
(d) 2π/3
Answer:
(d) 2π/3
Hint:
Let the amplitude of each wave be A and phase difference between them be φ. Then,

Question 73.
A sonometer wire is vibrating in the second overtone. In the wire there are ……
(a) two nodes and two antinodes
(b) one node and two antinodes
(c) four nodes and three antinodes
(d) three nodes and three antinodes
Answer:
(c) four nodes and three antinodes

Question 74.
If a resonance tube is sounded with a tuning fork of frequency 256 Hz, resonance occurs at 35 cm and 105 cm. The velocity of sound is about ……
(a) 358 m/s
(b) 512 m/s
(c) 524 m/s
(d) none of these
Answer:
(a) 358 m/s
Hint:

Question 75.
A wave of frequency 100 Hz is sent along a string towards a fixed end when this wave travels back after reflection, a node is formed at a distance of 10 cm from the fixed end of the string. The speed of the incident wave is ……
(a) 40 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
Answer:
(b) 20 m/s
Hint:
The fixed end is also a node distance between two nodes = \(\frac{\lambda}{2}\) = 10 cm
or λ = 20 cm = 0.2 cm
Speed v = fλ = 100 × 0.2 = 20 m/s

Question 76.
A standing wave is represented by y = A sin (100t) cos (0.01x) where y and A are in millimetres, t in seconds and x in metres. The velocity of the wave is ………
(a) 10⁴ m/s
(b) 1 m/s
(c) 10^-4 m/s
(d) not derivable from the above information
Answer:
(a) 10⁴ m/s

Question 77.
Two waves of the same frequency and intensity superimpose with each other in opposite phases. Then after superposition the ……
(a) intensity increases to four times
(b) intensity increase to two times
(c) frequency increases to four times
(d) none of the above
Answer:
(d) none of the above
Hint:
Since the waves are in opposite phases, the resultant intensity will be zero. The frequency remains the same. So, the correct choice is (d).

Question 78.
Two open organ pipes of lengths 50 cm and 50.5 cm produce 3 beats/s. Then the velocity of sound is …….
(a) 300 m/s
(b) 30 m/s
(c) 303 m/s
(d) 30.3 m/s
Answer:
(c) 303 m/s

Question 79.
If the ratio of the amplitudes of two waves is 4 : 3, then the ratio of maximum and minimum intensities is …….
(a) 16 : 9
(b) 49 : 16
(c) 7 : 1
(d) 49 :1
Answer:
(d) 49 : 1

Question 80.
An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 256 Hz, if the length of the column in centimeter is (velocity of sound in air = 340 m/s)
(a) 21.25
(b) 125
(c) 62.50
(d) 33.2
Answer:
(d) 33.2
Hint:

Question 81.
Two sound waves with wavelengths 5.0 cm and 5.5 cm, respectively each propagate in a gas with velocity 330 m/s. The number of beats per second will be ……..
(a) 0
(b) 1
(c) 6
(d) 12
Answer:
(c) 6
Hint:

Question 82.
Two vibrating tuning forks produce progressive waves given be y₁ =4 sin 500 πt and y₂ = 2 sin 506 πt where t is in seconds number of beats produced per minute is ……..
(a) 60
(b) 3
(c) 369
(d) 180
Answer:
(d) 180

Question 83.
The ratio of intensities of two waves is 16 : 9. If they produce interference, then the ratio of maximum and minium intensities will be ……..
(a) 4 : 3
(b) 49 : 1
(c) 64 : 27
(d) 81 : 49
Answer:
(b) 49 : 1

Question 84.
A closes organ pipe of length 20 cm is sounded with a tuning fork in resonance. What is the frequency of the tuning fork? (v = 332 m/s)
(a) 300 Hz
(b) 350 Hz
(c) 375 Hz
(d) 415 Hz
Answer:
(d) 415 Hz So,
Hint: In resonance, the frequency of the fork is equal to the frequency of the organ pipe,

Question 85.
In a resonance tube, the first resonance is obtained at 40 cm length, using a tuning fork of frequency 450 Hz. Ignoring end correction, the velocity of sound in air is
(a) 620 m/s
(b) 720 m/s
(c) 820 m/s
(d) 1020 m/s
Answer:
(b) 720 m/s

Question 86.
If we study the vibration of a pipe open at both ends, then which of the following statement is not true?
(a) open end will be antinode
(b) odd harmonics of the fundamental frequency will be generated
(c) all harmonics of the fundamental
(d) pressure change will be maximum at both ends.
Answer:
(d) pressure change will be maximum at both ends.
Hint:
Pressure change at open ends is zero.

Question 87.
The fundamental frequency of a closes organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is ……
(a) 80 cm
(b) 100 cm
(c) 120 cm
(d) 140 cm
Answer:
(c) 120 cm

Samacheer Kalvi 11th Physics Waves 2 Mark Questions

Question 1.
Define the term wave motion?
Answer:
Wave motion is a kind of disturbances which travels through a medium due to repeated vibrations of the particles of the medium about their mean positions, the disturbance being handed over from one particle to the next.

Question 2.
What is a progressive wave?
Answer:
A wave that travels from one point of the medium to another is called a progressive wave.

Question 3.
What is a plane progressive harmonic wave?
Answer:
If during the propagation of a wave through a medium the particles of the medium vibrate simple harmonically about their mean positions, than the wave is said to be plane progressive harmonic wave.

Question 4.
What do you mean by phase of a wave?
Answer:
The phase of a harmonic is a quantity that gives complete information of the wave at any time and at any position.

Question 5.
Define wave velocity or phase velocity?
Answer:
The distance covered by a wave in the direction of its propagation per unit time is called the wave velocity.

Question 6.
What are stationary waves?
Answer:
When two identical waves of same amplitude and frequency travelling in opposite directionals with the same speed along the same path superpose each other, the resultant wave does not travel in the either direction and is called stationary or standing waves.

Question 7.
What is meant by threshold of heating?
Answer:
The lowest intensity of sound that can be perceived by the human ear is called threshold of hearing. For a sound of frequency 10 kHz, the threshold of hearing is 10^-12 Wm^-2

Question 8.
What is meant by reverberation?
Answer:
The persistence of audible sound after the source has ceased to emit sound is called reverberation.

Question 9.
What is musical scale?
Answer:
A series of notes whose fundamental frequencies have definite ratios and which produce a pleasing effect on the ear when sounded in succession constitute a musical scale.

Question 10.
Define reverberation time?
Answer:
It is defined as the time which sound takes to fall in intensity to one millionth (10^-6) part of its original intensity after it was stopped.

Samacheer Kalvi 11th Physics Waves Numerical Problems

Question 1.
The fundamental frequency in an open organ pipe is equal to the 3rd harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm. What is the length of the open organ pipe.
Answer:

Question 2.
Two cars moving in opposite directions approach each other with speed of 22 ms^-1 and 16.5 ms^-1 respectively. The driver of the first car blows a horn having a frequency 400 Hz. To find the frequency heard by the driver of the second car.
Answer:

Question 3.
The second overtone of an open organ pipe has the same frequency as the 1st overtone of a closed pipe L metre long. Then what will be the length of the open pipe.
Answer:

The Length of the open pipe is two times of the length of the closed pipe.

Question 4.
A steel wire 0.72 m long has a mass of 5.0 × 10^-3 kg. If the wire is under a tension of 60 N. What is the speed of transverse waves on the wire?
Answer:

Question 5.
Estimate the speed of sound ¡n air at standard temperature and pressure by using
(i) Newton’s formula and
(ii) Laplace formula. The mass of 1 moIe of air = 29. 0 × 10^-3 kg. For air, γ = 1.4
Answer:

(ii) According to Newton’s formula speedof sound in air at S.T.P is

Question 6.
An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What Is the percentage increase in the apparent frequency?
Answer:
Here observer moves towards the stationary source.

The percentage increase in apparent frequency

Question 7.
Tube A has both ends open, while B has on end closed otherwise the two tubes are identical. What Is the ratio of fundamental frequency of the tubes A and B?
Answer:
The fundamental frequency for tube A with both ends open is f_A = \(\frac{v}{2 \mathrm{L}}\)
The fundamental frequency for tube B with one end closed is f_B = \(\frac{v}{4 \mathrm{L}}\)

Question 8.
A train moves towards a stationary observer with speed 34 mIs. The train sounds a whistle and its frequency registered by the observer is f₁. If the train’s speed is reduced to 17 m/s, the frequency registered f₂. If the speed of sound is 340 m/s, then find the ratio f₁/f₂
Answer:

Question 9.
A police car with a siren of frequency 8 kHz ¡s moving with uniform velocity 36 km/h towards a tall building which reflect the sound waves. The speed of sound In air is 320m/s. What is the frequency of the siren heard by the car driver?
Answer:
(a) Frequency received by the building.

The wall (source) reflect this frequency, So frequency heard by the car driver is

Question 10.
The displacement y of a wave travelling in the x-direction ¡s given by y = 10^-4 sin (600t – 2x + π/3)
Where x is expressed in metres and t is seconds. What is the speed of the wave motion (in ms^-1)?
Answer:

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