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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

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Samacheer Kalvi 11th Chemistry Chapter 1 Basic Concepts of Chemistry and Chemical Calculations Textual Evaluation Solved

I. Choose the Best Answer

Samacheer Kalvi 11th Chemistry Chapter 1 Solutions Question 1.
40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature is ………..
(a) 40 ml CO₂ gas
(b) 40 ml CO₂ gas and 80 ml H₂O gas
(c) 60 ml CO₂ gas and 60 ml H₂O gas
(d) 120 ml CO₂ gas
Answer:
(a) 40 ml CO₂ gas
Solution:
CH_4(g)  + 2O_2(g)  CO_2(g) + 2 H₂O_(l)

Content	CH4	O2	CO2
Stoichiometric coefficient	1	2	1
Volume of reactants allowed to react	40 mL	80 mL	–
Volume of reactant reacted and product formed	40 mL	80 mL	40 mL
Volume of gas after cooling to the room temperature	–	–	–

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct

Basic Concepts Of Chemistry And Chemical Calculations Question 2.
An element X has the following isotopic composition ²⁰⁰X = 90 %, ¹⁹⁹X = 8 % and ²⁰²X = 2 %. The weighted average atomic mass of the element X is closest to …………
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
Answer:
(d) 200 u
= \(\frac {(200 × 90) + (199 × 8) + (202 × 2) }{100}\) = 199.96 = 200 u

Basic Concepts Of Chemistry And Chemical Calculations Book Back Answers Question 3.
Assertion:
Two mole of glucose contains 12.044 × 10²³ molecules of glucose.

Reason:
Total number of entities present in one mole of any substance is equal to 6.02 × 10²²
(a) both assertion and reason are true and the 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:
(c) assertion is true but reason is false

Correct reason:
Total number of entities present in one mole of any substance is equal to 6.022 x 10²³

Samacheer Kalvi Guru 11th Chemistry Question 4.
Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) Both carbon and oxygen
(d) Neither carbon nor oxygen
Answer:
(b) Oxygen

Reaction 1:
2 C + O₂ → 2 CO₂
2 × 12 g carbon combines with 32 g of oxygen.
Hence, Equivalent mass of carbon = \(\frac {2 × 12}{32}\) × 8 = 6

Reaction 2:
C + O₂ → 2 CO₂
12 g carbon combines with 32 g of oxygen.
Hence, Equivalent mass of carbon = \(\frac {12}{32}\) × 8 = 3

11th Chemistry Samacheer Kalvi Question 5.
The equivalent mass of a trivalent metal element is 9 g eq^-1 the molar mass of its anhydrous oxide is ………..
(a) 102 g
(b) 27 g
(c) 270 g
(d) 78 g
Answer:
(a) 102 g
Let the trivalent metal be M³⁺
Equivalent mass = mass of the metal / valance factor
9g eq^-1 = mass of the metal / 3 eq
Mass of the metal = 27 g
Oxide formed M₂O₃
Mass of the oxide = (2 × 27) + (3 × 16) = 102 g

11th Chemistry Lesson 1 Question 6.
The number of water molecules in a drop of water weighing 0.018 g is ………….
(a) 6.022 × 10²⁶
(b) 6.022 × 10²³
(c) 6.022 × 10²⁰
(d) 99 × 10²²
Answer:
(c) 6.022 × 10²⁰
Weight of the water drop = 0.0 18 g
No. of moles of water in the drop = Mass of water / molar mass = 0.018/18 = 10^-3 mole
No of water molecules present ¡n I mole of water = 6.022 × 10²³
“No. water molecules in one drop of water (10 mole) = 6.022 × 10²³ × 10^-3
= 6.022 × 10²⁰

Chemistry Class 11 Samacheer Kalvi Question 7.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is ………..
(a) 0 %
(b) 4.4 %
(c) 16 %
(d) 8.4 %
Answer:
(c) 16%
Mg CO₃ → MgO + CO₂↑
Mg CO₃ : (1 × 24) + (1 × 12) + (3 × 16) = 84 g
CO₂ : (1 × 12) + (2 × 16) 44g
100% pure 84 g MgCO₃ on heating gives 44 g CO₂
Given that I g of MgCO₃ on heating gives 0.44 g CO₂
Therefore, 84 g MgCO₃ sample on heating gives 36.96 g CO₂ = 100%
Percentage of purity of the sample = \(\frac{100 \%}{44 \mathrm{g} \mathrm{CO}_{2}}\) × 36.96 g CO₂ = 84%
Percentage of impurity = 16%

Samacheer Kalvi 11th Chemistry Question 8.
When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is –
(a) 3
(b) 0.75
(c) 0.075
(a) 0.3
Answer:
(c) 0.075

The amount of CO₂ released, x = 3.3 g
No. of moles of CO₂ released = 3.3 / 44 = 0.075 mol

Samacheer Kalvi 11th Chemistry Solutions Question 9.
When 22.4 liters of H₂ (g) is mixed with 11.2 liters of Cl₂ (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to ………..
(a) 2 moles of HCl (g)
(b) 0.5 moles of HCl (g)
(c) 1.5 moles of HCl (g)
(d) 1 moles of HCl (g)
Answer:
(d) 1 moles of HCl (g)
Solution:
H₂(g) + Cl₂(g) → 2 HCl (g)

Content	H2(g)	cl2(g)	HCl (g)
Stoichiometric coefficient	1	1	2
No. of moles of reactants allowed to react at 273 K and 1 atm pressure	22.4 L (1 mol)	11.2 L (0.5 mol)	—
No. of moles of reactant reacted and product formed	0.5	0.5	1
A mount of HCl formed 1 mol

11th Chemistry Chapter 1 Solutions Question 10.
Hot concentrated sulfuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior?
(a) Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
(b) C + 2H₂SO₄ → 4 CO₂ + 2SO₂ + 2H₂O
(c) BaCl₂ + H₂SO₄ → BaSO₄+ 2HCl
(d) none of the above
Answer:
(c) BaCl₂ + H₂SO₄ → BaSO₄+ 2HCl

11th Chemistry Chapter 1 Book Back Answers Question 11.
Choose the disproportional reaction among the following redox reactions.
(a) 3Mg (s) + N₂(g) → Mg₂N₂ (s)
(b) P₄ (s) + 3NaOH + 3H₂O → PH₃(g) + 3NaH₂PO₂ (aq)
(c) Cl₂ (g) + 2Kl (aq) → 2KC1 (aq) + I₂
(d) Cr₂O₃ (s) + 2Al (s) → A₂O₃ (s) + 2Cr (s)
Answer:
(b) P₄ (s) + 3NaOH + 3H₂O → PH₃(g) + 3NaH₂PO₂ (aq)

Basic Concepts Of Chemistry And Chemical Calculations Pdf Question 12.
The equivalent mass of potassium permanganate in alkaline medium is
MnO₄ + 2H₂O + 3e^– → MnO₂ + 4OH^–
(a) 31.6
(b) 52.7
(c) 79
(d) None of these
Answer:
(b) 52.7
The reduction reaction of the oxidizing agent(MnO₄) involves gain of 3 electrons.
Hence the equivalent mass = (Molar mass of KMnO₄) / 3 = 158.1 / 3 = 52.7

11th Chemistry 1st Chapter Question 13.
Which one of the following represents 180 g of water?
(a) 5 Moles of water
(b) 90 moles of water
(c) \(\frac{6.022 \times 10^{23}}{180}\) Molecules of water
(d) 6.022 × 10²⁴ Molecules of water
Answer:
(d) 6.022 x 10²⁴ Molecules of water
No. of moles of water present in 180 g
= Mass of water / Molar mass of water
= 180 g /18 g mol^-1 = 10 moles
One mole of water contains
= 6.022 × 10²³ water molecules
10 mole of water contains = 6.022 × 10²³ × 10
= 6.022 × 10²⁴ water molecules

Class 11 Chemistry All Formulas Question 14.
7.5 g of a gas occupies a volume of 5.6 liters at 0°C and 1 atm pressure. The gas is …………
(a) NO
(b) N₂O
(c) CO
(d) CO₂
Answer:
(a) NO
7.5 g of gas occupies a volume of 5.6 liters at 273 K and 1 atm pressure Therefore, the mass of gas that occupies a volume of 22.4 liters –
\(\frac {7.5 g}{5.6 L}\) × 22. 4 L = 30g
Molar mass of NO (14 + 16) = 30g

11th Chemistry Solutions Samacheer Kalvi Question 15.
Total number of electrons present in 1.7 g of ammonia is ………..
(a) 6.022 × 10²³
(b) \(\frac{6.022 \times 10^{22}}{1.7}\)
(c) \(\frac{6.022 \times 10^{24}}{1.7}\)
(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Answer:
(a) 6.022 × 10²³
No. of electrons present in one ammonia (NH₃) molecule (7 + 3) = 10
No. of moles of ammonia = \(\frac {Mass}{Molar mass}\)
= \(\frac{1.7 \mathrm{g}}{17 \mathrm{g} \mathrm{mol}^{-1}}\) = 0.1 mol
No. of molecules present in 0ne ammonia
= 0.1 × 6.022 × 10²³ = 6.O22 × 10²²
No. of electrons present in 0.1 mol of ammonia
10× 6.022 × 10²² = 6.022 × 10²³

Samacheer Kalvi Class 11 Chemistry Solutions Question 16.
The correct increasing order of the oxidation state of sulphur in the anions SO₄^2-, SO₃^2-, S₂O₄^2-,S₂O₆^2- is ………..
(a) SO₃^2- < SO₃^2- < S₂O₄^2- < S₂O₆^2-
(b) SO₄^2- < S₂O₄^2- < S₂O₆^2-<SO₃^2-
(c) S₂O₄^2- < SO₃^2- < S₂O₆^2- < SO₄^2-
(d) S₂O₆^2- < S₂O₄^2- < SO₄^2- < SO₃^2-
Answer:
(c) S₂O₄^2- < SO₃^2- < S₂O₆^2- < SO₄^2-

Class 11 Chemistry Solutions Samacheer Kalvi Question 17.
The equivalent mass of ferrous oxalate is ……….

Answer:

Samacheer Kalvi 11 Chemistry Solutions Question 18.
If Avogadro number were changed from 6.022 × 10²³ to 6.022 × 10²⁰, this would change ………..
(a) the ratio of chemical species to each other in a balanced equation
(b) the ratio of elements to each other in a compound
(c) the definition of mass in units of grams
(d) the mass of one mole of carbon
Answer:
(d) the mass of one mole of carbon

Samacheerkalvi.Guru 11th Chemistry Question 19.
Two 22.4 liter containers A and B contains 8 g of O₂ and 8 g of SO₂ respectively, at 273 K and 1 atm pressure, then ……….
(a) number of molecules in A and B are same
(b) number of molecules in B is more than that in A
(c) the ratio between the number of molecules in A to the number of molecules in B is 2 : 1
(d) number of molecules in B is three times greater than the number of molecules in A
Answer:
(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1

Question 20.
What is the mass of precipitate formed when 50 ml of 8.5% solution of Ag NO₃ is mixed with 100 ml of 1.865% potassium chloride solution?
(a) 3.59 g
(b) 7 g
(c) 14 g
(d) 28 g
Answer:
(a) 3.59 g
AgNO₃ + KCl → KNO₃ + AgCl
Solution:
50 mL of 8.5% solution contains 4.25 g of AgNO₃
No. of moles of AgNO₃ present in 50 mL of 8.5% AgNO₃ solution
= Mass / Molar mass = 4.25 / 170 = 0.025 moles
Similarly, No of moles of KCl present in loo mL of 1.865% KCl solution
= 1.865 / 74.5 = 0.025 moles
So total amount of AgCl formed is 0.025 moles (based on the stoichiometry calculator)
Amount of AgCl present in 0.025 moles of AgCl
= no. of moles × molar mass
= 0.025 × 143.5 = 3.59 g

Question 21.
The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25°C and 1 atm pressure) is 1.1g. The molar mass of the gas is ………..
(a) 66.25 g mol^-1
(b) 44 g mol^-1
(c) 24.5 g mol ^-1
(d) 662.5 g mol^-1
Answer:
(b) 44 g mol^-1
Solution:
No. of moles of a gas that occupies a volume of 6 12.5 ml at room temperature and pressure
(25° C and 1 atm pressure)
= 612.5 × 10^-3 L/24.5 L mol^-1
= 0.02 5 moles
We know that,
Molar mass = Mass / no. of moles
= 1.1 g/0.025 mol = 44 g mol^-1

Question 22.
Which of the following contain same number of carbon atoms as in 6 g of carbon -12?
(a) 7.5 g ethane
(b) 8 g methane
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)
Solution:
No. of moles of carbon present in 6 g of C – 12 = Mass / Molar mass
= 6/12 = 0.5 moles = 0.5 × 6.022 × 10²³ carbon atoms.
No. of moles in 8 g of methane = 8 116 = 0.5 moles
= 0.5 × 6.022 × 10²³ carbon atoms.
No. of moles in 7.5 g of ethane = 7.5 / 16 = 0.25 moles
= 2 × 0.25 × 6.022 × 10²³ carbon atoms.

Question 23.
Which of the following compound( s) has/have percentage of carbon same as that in ethylene (C₂H₄)?
(a) propene
(b) ethyne
(c) benzene
(d) ethane
Answer:
(a) propene
Solution:
Molar mass of carbon
Percentage of carbon in ethylene(C₂H₆) =
= \(\frac {24}{28}\) × 100 = 85.71%
Percentage of carbon in propene (C₃H₆) = \(\frac {24}{28}\) × 100 = 85.7 1%

Question 24.
Which of the following is/are true with respect to carbon – 12?
(a) relative atomic mass is 12 u
(b) oxidation number of carbon is +4 in all its compounds.
(c) I mole of carbon -12 contain 6.022 × 10²² carbon atoms.
(d) all of these
Answer:
(a) relative atomic mass is 12 u

Question 25.
Which one of the following is used as a standard for atomic mass?
(a) ₆C¹²
(b) ₇C¹²
(c) ₆C¹³
(d) ₆C¹⁴
Answer:
(a) ₆C¹²

II. Write brief answer to the following questions

Question 26.
Define relative atomic mass.

On the basis of carbon, the relative atomic mass of element is defined as the ratio of mass of one atom of the element to the mass of l/12th mass of one atom of Carbon – 12.

Question 27.
What do you understand by the term mole?
Answer:
The mole is defined as the amount of a substance which contains 6.023 x 10²³ particles such as atoms, molecules or ions. It is represented by the symbol

Question 28.
Define equivalent mass.
Answer:
The equivalent mass of an element is the number of parts of the mass of an element which combines with or displaces 1.008 parts of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine.

Question 29.
What do you understand by the term oxidation number?
Answer:
Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely. The oxidation numbers reflect the number of electrons “transferred”.

Question 30.
Distinguish between oxidation and reduction.
Answer:
Oxidation:
According to the classical concept, oxidation is a process of addition of oxygen or removal of hydrogen.
Removal of hydrogen
2H₂S + O₂ → H₂O + 2S
Addition of oxygen
C + O₂ → CO₂
According to the electronic concept, loss of electrons is called oxidation reaction.
Ca → Ca²⁺ + 2e^–
During oxidation, oxidation number increases.
Dining oxidation, reducing agent gets oxidised.

Reduction:
Reduction is a process of removal of oxygen or addition of hydrogen.
Addition of hydrogen
Ca + H₂ → CaH₂
Removal of oxygen
Zn O + C → Zn + CO
According to the electronic concept, gain of electrons is called reduction reaction.
Zn²⁺ + 2e^– → Zn
During reduction, oxidation number decreases.
During reduction, oxidising agent gets reduced.

Question 31.
Calculate the molar mass of the following compounds.

1. urea [CO(NH₂)₂]
   
2. acetone [CH₃COCH₃]
   
3. boric acid [H₃BO₃]
4. sulphuric acid [H₂SO₄]

Answer:
1. urea [CO(NH₂)₂]
Atomic mass of C =12
Atomic mass of O =16
Atomic mass of 2(N) = 28
Atomic mass of 4(H) = 4
∴ Molar mass of Urea = 60

2. Acetone [CH₃COCH₃]
Atomic mass of 3(C) = 36
Atomic mass of 1(0) = 16
Atomic mass of 6(H) = 6
∴ Molar mass of Acetone = 58

3. Boric acid [H₃BO₃]
Atomic mass of B = 10
Atomic mass of 3(H) = 3
Atomic mass of 3(O) = 48
∴ Molar mass of Boric acid = 61

4. Sulphuric acid ₂[H₂SO₄]
Atomic mass of 2(H) = 2
Atomic mass of 1(S) = 32
Atomic mass of 4(O) = 64
∴ Molar mass of Sulphuric acid = 98

Question 32.
The density of carbon dioxide is equal to 1.977 kg m^-3 at 273 K and 1 atm pressure. Calculate the molar mass of CO₂
Answer:
Molecular mass = Density x Molar volume
Molar volume of CO₂ = 2.24 x 10^-2 m³
Density of CO₂ = 1.977 kg m^-3
Molecular mass of CO₂ = 1.977 x 10³ gm^-3  x 2.24 x 10^-2 m³
= 1.977 × 10^-1 × 2.24 = 44 g

Question 33.
Which contains the greatest number of moles of oxygen atoms?

1. 1 mol of ethanol
2. 1 mol of formic acid
3. 1 mol of H₂O

Answer:
1. 1 mol of ethanol
C₂H₅OH (ethanol) – Molar mass = 24 + 6 + 16 = 46
46 g of ethanol contains 1 × 6.023 × 10²³ number of oxygen atoms.

2. 1 mol of formic acid.
HCOOH (formic acid) – Molar mass = 2+12 + 32 = 46
46 g of HCOOH contains 2 × 6.023 × 10²³ number of oxygen atoms.

3. 1 mol of H₂O
H₂O (water) – Molar mass = 2 + 16 = 18
18 g of water contains 1 × 6.023 × 10²³ number of oxygen atoms.
∴ 1 mole of formic acid contains the greatest number of oxygen atoms.

Question 34.
Calculate the average atomic mass of naturally occurring magnesium using the following data

Isotope	Isotopic atomic mass	Abundance (%)
Mg24	23.99	78.99
Mg26	24.99	10
Mg25	25.98	11.01

Answer:
Isotopes of Mg.
Atomic mass = Mg²⁴ = 23.99 x \(\frac {783. 99}{100}\) = 18.95
Atomic mass = Mg²⁶ = 24.99 x \(\frac {10}{100}\) = 2.499
Atomic mass = Mg²⁵ = 25.98 x \(\frac {11.01}{100}\) = 2.860
Average Atomic mass = 24.309
Average atomic mass of Mg = 24.309

Question 35.
In a reaction x + y + z₂ → xyz₂, identify the limiting reagent if any, in the following reaction mixtures.
(a) 200 atoms of x + 200 atoms of y + 50 molecules of z₂
(b) 1 mol of x + 1 mol of y + 3 mol of z₂
(c) 50 atoms of x + 25 atoms of y + 50 molecules of z₂
(d) 2.5 mol of x + 5 mol of y + 5 mol of z₂
Answer:
x + y + z₂
(a) 200 atoms of x + 200 atoms of y + 50 molecules of z₂ According to the reaction, 1 atom of x reacts with one atom of y and one molecule of z to give product. In the case (a) 200 atoms of x, 200 atoms of y react with 50 molecules of z₂ (4 part) i.e. 50 molecules of z₂ react with 50 atoms of x and 50 atoms of y. Hence z is the limiting reagent.

(b) 1 mol of x + 1 mol of y + 3 mol of z₂
According to the equation 1 mole of z₂ only react with one mole of x and one mole of y. If 3 moles of z₂ are there, z is limiting reagent.

(c) 50 atoms of x + 25 atoms of y + 50 molecules of z₂
25 atoms of y react with 25 atoms of x and 25 molecules of z₂. So y is the limiting reagent.

(d) 2.5 mol of x + 5 mol of y + 5 mol of z₂
2.5 mol of x react with 2.5 mole of y and 2.5 mole of z₂. So x is the limiting reagent.

Question 36.
Mass of one atom of an element is 6.645 × 10^-23 g. How many moles of element are there in 0.320 kg?
Answer:
Mass of one atom of an element = 6.645 × 10^-23 g = Atomic mass.
Mass of given element = 0.320 kg
Number of moles =
Atomic mass

= 48.156 x 10^-23
= 4.8156 x 10^-24 moles.

Question 37.
What is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
Answer:
Molecular mass:

• Relative molecular mass is defined as the ratio of the mass of the molecule to the unified atomic mass unit.
•  It can be calculated by adding the relative atomic masses of its constituent atoms.
• For carbon monoxide (CO) Molecular mass = Atomic mass of carbon + Atomic mass of oxygen 12 + 16 = 28 u.

Molar mass:

• It is defined as the mass of one mole of a substance.
• The molar mass of a compound is equal to the sum of the relative atomic masses of its constituent expressed in g mol^-1.
•  For carbon monoxide (CO) 12 + 16 = 28 g mol^-1 Both molecular mass and molar mass are numerically same but the units are different.

Question 38.
What is the empirical formula of the following?

1. Fructose (C₆H₁₂O₆) found in honey
2. Caffeine (C₈H₁₀N₄O₂) a substance found in tea and coffee.

Answer:
1. Fructose (C₆H₁₂O₆)
Empirical formula is the simplest formula. So it is divided by 6 and so empirical formula is CH₂O.

2. Caffeine (C₈H₁₀N₄O₂)
Simplified formula = \(\frac {molecular formula}{2}\)
Empirical formula = C₄H₅N₂O.

Question 39.
The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of AC = 21 u Atomic mass of 0 = 16 u) 2Al + Fe₂O₂ → Al₂O₃ + 2Fe; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide.

1. Calculate the mass of Al₂O₃ formed.
2. How much of the excess reagent is left at the end of the reaction?

Answer:

1. As per balanced equation 54 g A1 is required for 112 g of iron and 102 g of Al₂O₃.
54 g of Al gives 102 g of Al₂O₃.
∴ 324 g of Al will give \(\frac{102}{54}\) x 324 = 612 g of Al₂O₃.

2. 54 g of Al requires 160 g of Fe₂O₃ for welding reaction.
∴ 324 g of Al will require \(\frac {160}{54}\) x 324 = 960 g of Fe₂O₃.
∴ Excess Fe₂O₃ – Un reacted Fe₂O₃ = 1120 – 960 = 160 g
160 g of excess reagent is left at the end of the reaction.

Question 40.
How many moles of ethane is required to produce 44 g of CO₂ (g) after combustion.
Answer:

∴ 44g of CO₂ = I mole of CO₂
2 moles of CO₂ is produced by 1 mole of ethane.
∴ 1 mole of CO₂ will be produced by = ?
∴ To produce 1 mole of CO₂, the required mole of ethane is = \(\frac {1}{2}\) x 1 = 0.5 mole of ethane.

Question 41.
Hydrogen peroxide is an oxidizing agent. It dioxides ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
Answer:
H₂O₂ – Oxidizing agent
Fe²⁺ + H₂O₂ → Fe³⁺ + H₂O (Acetic medium)
Ferrous ion is oxidized by H₂O₂ to Ferric ion.
The balanced equation is Fe²⁺ → Fe³⁺ + e^– x 2

Question 42.
Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
Answer:

Empirical formula = C₆H₆O
Va-pour density 47
∴ Molecular mass = 2 x vapor density = 2 x 47 = 94
Molecular formula Empirical formula x n
Molecular mass x n
n = \(\frac { Molecular mass }{ Empirical formula mass }\) = \(\frac {94}{94}\) = 1
∴ Molecular formula = C₆H₆O

Question 43.
A Compound on analysis gave Na = 14.31% S = 9.97% H = 6.22% and O = 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization, (molecular mass of the compound is 322).
Answer:

All H combines with 10 oxygen atoms to form as 10H₂O.
So the empirical formula is Na₂SO₄ .10H₂ O
Empirical formula mass = (23 x 2) + (32 x 1) + (16 x 4) + (10 x 18)
= 46 + 32 + 64 + 180 = 322
n = \(\frac { Molecular mass }{ Empirical formula mass }\) = \(\frac {322}{322}\) = 1
Molecular formula = Na₂SO₄. 10H₂O

Question 44.
Balance the following equations by oxidation number method

1. K₂Cr₂ O₇ + KI + H₂SO₂ → K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
2. KMnO₄ + Na₂SO₃ → Mn0₂ + Na₂SO₄ + KOH
3. Cu+ HNO₃ → Cu(N0₃)₂ + NO₂ + H₂O
4. H₂C₂O₄ + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + CO₂ + H₂O

Answer:
1. K₂Cr₂ O₇ + KI + H₂SO₂ → K₂SO₂ + Cr₂(SO₄)₃ + I₂ + H₂O
Step – 1.

Step – 2
K₂Cr₂ O₇ + 6KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + H₂O
Step – 3
To balance other atoms
K₂Cr₂ O₇ + 6KI + H₂SO₄ → 4K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + H₂O
Step – 4
K₂Cr₂ O₇ + 6KI + 7H₂SO₄ → 4K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + 7H₂O

2. KMnO₄ + Na₂SO₃ → MnO₂ + Na₂SO₄ + KOH (Alkaline medium)
Step – 1

Step – 2
2KMnO₄ + 3Na₂SO₃ → 2MnO₂ + 3Na₂SO₄ + KOH
Step – 3
balancing potassium, KOH is multiplied by 2
2KMnO₄ + 3Na₂SO₃ → 2MnO₂ + 3Na₂SO₄ + KOH
Step – 4
To balance H atom, H₂0 is added on reactant side.
2KMnO₄ + 3Na₂SO₃ + H₂O → 2MnO₂ + 3Na₂SO₄ + KOH

3. Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O
Step – 1

Step – 2
Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O
Step – 3
To balance Nitrogen, 2HNO₃ is multiplied by 2 and NO₂ is multiplied by 2
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + H₂O
Step 4.
To balance oxygen, H₂O is multiplied by 2
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

4. H₂C₂O₄ + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + CO₂ + H₂O
Step – 1

Step – 2
5 H₂C₂O₄ + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + 10 CO₂ + H₂O
Step – 3
To balance K, KMnO₄ and MnSO₄ are multiplied by 2
5 H₂C₂O₄ + 2KMnO₄ + H₂SO₄ → K₂SO₄ + 2MnSO₄ + 10 CO₂ + H₂O
Step – 4
To balance O and H, H₂O and H₂SO₄ are multiplied by 3 and 6.
5 H₂C₂O₄ + 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 10 CO₂ + 8H₂O

Question 45.
Balance the following equations by ion electron method.

1. KMnO₄ + SnCl₂  + HCl → MnCl₂  + SnCl₄  + H₂O + KCl
2. C₂O₄^2- + Cr₂ O_7 2- → Cr ³⁺ + CO₂  (in acid medium)
3. Na₂S₂O₃ + I₂ → Na₂S₄O₆ + NaI (in acid medium)
4. Zn + NO₃^– → Zn²⁺ + NO

Answer:
1. KMnO₄ + SnCl₂ + HCl → MnCl₃ + SnCl₃ +H₂O + KCl
Oxidation half reaction: (loss of electrons)

Reduction half reaction: (gain of electrons)

Add H₂O to balance oxygen atoms.

Add HCl to balance hydrogen atoms
KMnO₄ + 5e_– + 8HCl → MnCl₂ + 4H₂O ………(4)
To equalize the number of electrons equation (1) x 5 and equation (2) x 2
5SnCl₂ → 5SnCl₄ + 10e^–
2KMnO₄ + 16HCl + 10e^– → 2MnCl₂ + 4H₂O + 2KCl
2KMnO₄ + 5SnCl₂ + 16HCl → 5SnCl₄ + 2MnCl₂ + 4H₂O + 2KCl

2. C₂O₄^2- + Cr₂ O₇^2- → Cr ³⁺ + CO₂  (in acid medium)
Oxidation half reaction:

Reduction half reaction:

To balance oxygen atoms, H₂O is added on RHS of equation (2)
Cr₂O₇^2- + 6e^– → 2Cr³⁺ + 7 H₂O ……….(3)
To balance Hydrogen atoms, H⁺ is added on LHS of equation (1)
C₂O₄^2- + 14H⁺ → 2CO₂ + 2e^– ……..(4)
To equalize the number of electrons gained and lost, multiply the equation (4) x 3.

3. Na₂S₂O₃ + I₂ → Na₂S₄O₆ + NaI (in acid medium)
Oxidation half reaction: (Loss of electron)
Na₂S₂O₃ → Na₂S₄O₆ + 2e^2- ……..(1)
Reduction half reaction: (Gain of electron)
I₂ + 2e^2-→ 2NaI …………(2)
Adding (1) and (2)

To balance oxygen,
2Na₂S₂O₃ + I₂ → Na₂S₂O₂ + 2NaI

In acidic medium
4. Zn + NO₃^– → Zn²⁺ + NO
Half reactions are –

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations In Text Questions – Evaluate Yourself

Question 1.
By applying the knowledge of chemical classification, classify each of the following Into elements, compounds or mixtures.

1. Sugar
2. Sea water
3. Distilled water
4. Carbon dioxide
5. Copper wire
6. Table salt
7. Silver plate
8. Naphthalene balls

Answer:

Elements	Compounds	Mixtures
Copper wire (Cu) Silver plate (Ag)	Sugar Distilled water Carbon dioxide Naphthalene balls	Sea water Table salt

Question 2.
Calculate the molar mass of the following.

1. Ethanol (C₂H₅OH)
2. Potassium permanganate (KMnO₄)
3. Potassium dichromate (K₂Cr₂O₇)
4. Sucrose (C₁₂H₂₂O₁₁)

Answer:
(1) Ethanol (C₂H₅OH)
Molar mass = (2 × 12) + (6 × 1) + (1 × 16)
= 24 + 6+16 = 46 g mol^-1

(2) Potassium permanganate (KMnO₄)
Molar mass = 39 + 55 + (4 × 16)
= 39 + 55 + 64 = 158 g mol^-1

(3) Potassium dichromate (K₂Cr₂O₇)
Molar mass = (39 × 2) + (2 × 52) + (7 × 16)
= 78 + 104 + 112 = 294 g mol^-1

(4) Sucrose (C₁₂H₂₂O₁₁)
Molar mass = (12 × 12) + (22 × 1) + (11 × 16) = 144 + 22 + 176 = 342 g mol^-1

Question 3.
(a) Calculate the number of moles present in 9 g of ethane.
Answer:
Mass of ethane = 9 g
Molar mass of ethane C₂H₆ = 30 g mol^-1
No. of moles = \(\frac {Mass}{Molar mass}\) = \(\frac {9}{30}\) = 0.3 mol.

(b) Calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 K and 3 atm pressure.
Answer:
Molar volume of oxygen = 22400 ml.
22400 ml of oxygen contains 6.023 x 10²³ molecules.
224 ml of oxygen contain \(\frac{6.023 \times 10^{23}}{22400}\) x 224
\(\frac{6.023 \times 10^{23}}{100}\) = 6.023 × 10²¹

Question 4.
(a) 0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal.
Answer:
Mass of the metal = W₁ = 0.456 g.
Mass of the metal chloride = W₂ = 0.606 g.
∴Mass of chlorine = W₂ – W₁ = 0.606 – 0.456 = 0.15 g
0.15 g of chlorine combine with 0.456 g of metal.
∴ 35.46 g of chlorine will combine with \(\frac {0.456}{0.15}\) x 35.46 = 107.79 g eq^-1

(b) Calculate the equivalent mass of potassium dichromate. The reduction half – reaction in acid medium is –
Cr₂O₇^2- + 14H⁺ +6e^– → 2 Cr³⁺ + 7H₂O
Answer:
Cr₂O₇^2- + 14H⁺ +6e^– → 2 Cr³⁺ + 7H₂O

Molar mass of K₂Cr₂O₇ = 294.18
∴ Equivalent mass of K₂Cr₂O₇ = \(\frac { 294.18}{6}\) = 49.03

Question 5.
A Compound on analysis gave the following percentage composition C = 54.55%, H = 9.09%, O = 36.36%. Determine the empirical formula of the compound.
Answer:

Empirical formula = C₂H₄O

Question 6.
Experimental analysis of a compound containing the elements x,y,z on analysis gave the following data, x = 32 %, y = 24 %, z = 44 %. The relative number of atoms of x, y and z are 2,1 and 0.5, respectively. (Molecular mass of the compound is 400 g) Find out.

1. The atomic masses of the element x,y,z.
2. Empirical formula of the compound and
3. Molecular formula of the compound.

Answer:
Element x = 32% ; y = 24% ; z = 44%
Relative number of atoms x = 2 ; y = 1 ; z = 0.5
Molar mass of the compound = 400 g.

1. Atomic mass of the element.
Relative number of atoms
∴  Atomic mass
Atomic mass x = \(\frac {32}{2}\) = 16
Atomic mass y = \(\frac {24}{1}\) = 24
Atomic mass z = \(\frac {44}{0.5}\) = 88

2. Empirical formula of the compound x₄ y₂ Z₁
Molecular mass of the compound = 400
= \(\frac {400}{200}\) = 2

3. Molecular formula of the compound = x₈ y₄ z₂

Question 7.
The balanced equation for a reaction is given below 2x + 3y → 41 + m When 8 moles of x react with 15 moles of y, then –

1. Which is the limiting reagent?
2. Calculate the amount of products formed.
3. Calculate the amount of excess reactant left at the end of the reaction.

Answer:
2 x + 3 y → 41 + m

1. 2x reacts with 3y to give products.
5x reacts with l5y means, y is the excess because 8 moles of x should react withn 4 x 3y = 12y moles of y to give products. In this reaction 15y moles are used. Therefore, 3 moles of y is excess and it is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 x 41 i.e. 161 and 4m as product.
8x + 12y → 161 + 4m

3.  At the end of the reaction, the excess reactant left in 3 moles of y.

Question 8.
Balance the following equation using oxidation number method
AS₂ S₃  + HNO₃ + H₂O → H₃ASO₄ + H₂SO₄ + NO
Answer:

To balance oxygen and sulphur, H₂O and H₂SO₄ are added.
3AS₂S₃ + 2HNO₃ + H₂O → 6H₃ASO₄ + 2NO + H₂SO₂
3AS₂S₃ + 28HNO₃ + 4H₂O → 6H₃ASO₄ + 28NO + 9H₂SO₄

Samacheer Kalvi 11th Chemistry Solutions Basic Concepts of Chemistry and Chemical Calculations Textual Calculations based on Stolchlometry Solved

Question 1.
How many moles of hydrogen is required to produce 10 moles of ammonia?
Answer:
The balanced stoichiometric equation fòr the formation of ammonia is
N₂(g) + 3H₂(g) → 2NH₂(g)
4s per the stoichiometric equation, to produce 2 moles of ammonia, 3 moles of hydrogen are required.
∴ to produce 10 moles of ammonia,

= 15 moles of hydrogen are required.

Question 2.
Calculate the amount of water produced by the combustion of 32 g of methane.
Answer:
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g)
As per the stoichiometric equation,
Combustion of 1 mole (16 g) CH₄ produces 2 moles (2 x 18 = 36 g) of water.

Combustion of 32 g CH₄ produces

Question 3.
How much volume of carbon dioxide is produced when 50 g of calcium carbonate is heated completely under standard conditions?
Answer:
The balanced chemical equation is,

As per the stoichiometric equation,
1 mole (100g) CaCO₃ on heating produces 1 mole CO₂

At STP, 1 mole of CO₂ occupies a volume of 22.7 litres
At STP, 50 g of CaCO₃ on heating produces,

= 11.35 litres of CO₂

Question 4.
How much volume of chlorine is required to form 11.2 L of HCl at 273 K and 1 atm pressure ?
Answer:
The balanced equation for the formation of HCl is,
H₂ (g) + Cl₂ (g) → 2 HCl (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCl, 1 mole of chlorine gas is required.
To produce 44.8 liters of HCl, 22.4 liters of chlorine gas are required
∴ To produce 11.2 liters of HCl,

= 5.6 litres of chlorine are required.

Question 5.
Calculate the percentage composition of the elements present in magnesium carbonate. How many kilogram of CO₂ can be obtained by heating 1 kg of 90 % pure magnesium carbonate.
Answer:
The balanced chemical equation is

Molar mass of MgCO₃ is 84 g mol^-1.
84 g MgCO₃ contain 24 g of Magnesium.
∴ 100 MgCO₃ contain

= 28.57 g Mg i.e. percentage of magnesium = 28.57 %.
84 g MgCO₃ contain 12 g of carbon


= 14.29 g of carbon
Percentage of carbon = 14.29 %.
84 g MgCO₃ contain 48 g of oxygen
∴ 100 g MgCO₃ contains

= 57.14 g of oxygen.
∴ Percentage of oxygen = 57.14 %.
As per the stoichiometric equation,
84 g of 100 % pure Mg CO₃ on heating gives 44 g of CO₂.
∴1000 g of 90 % pure Mg CO₃ gives
\(\frac {44 g}{84 g x 100 %}\) x 90 % x 1000 g
= 471.43 g of CO₂
= 0. 471 kg of CO₂

Question 6.

(1) If the entire quantity of all the reactants is not consumed in the reaction which is the limiting reagent ?
(2) Calculate the quantity of urea formed and un reacted quantity of the excess reagent. The balanced equation is

Answer:
(1) The entire quantity of ammonia is consumed in the reaction. So ammonia is the limiting reagent. Some quantity of CO₂ remains unreacted, so CO₂ is the excess reagent.

(2) Quantity of urea formed = number of moles of urea formed x molar mass of urea
= 19 moles x 60 g mol^-1
= 1140 g = 1.14 kg
Excess reagent leftover at the end of the reaction is carbon dioxide. Amount of carbon dioxide leftover = number of moles of CO₂ left over x molar mass of CO₂
= 7 moles x 44 g mol^-1 = 308 g.

Samacheer Kalvi 11th Chemistry Solutions Basic Concepts of Chemistry and Chemical Calculations Additional Questions Solved

I. Choose the correct answer from the following

Question 1.
Which one of the following is the standard for atomic mass?
(a) ₁H¹
(b) ₆6C¹²
(c) ₆C¹⁴
(d) ₈O¹⁶
Answer:
(b) ₆6C¹²
Hint:
Standard element used to determine atomic mass is 6 Cl₂

Question 2.
One mole of CO₂ contains ………….
(a) 6.023 x 10²³ atoms of C
(b) 6.023 x 10²³ atoms of O
(c) 18.1 x 10²³ molecules of CO₂
(d) 3g atoms of CO₂
Answer:
(a) 6.023 x 10²³ atoms of C
Hint:
One mole of any substance contains Avogadro number of atoms. In this carbon one mole is present and oxygen two atoms are present. So, 6.023 x 10²³ atoms of C is correct.

Question 3.
The largest number of molecules is in
(a) 54 g of nitrogen pent oxide
(b) 28 g of carbon dioxide
(c) 36 g of water
(d) 46 g of ethyl alcohol
Answer:
(c) 36 g of water
Hint:(a) 54 g of N₂O₅
N₂O₅ = Molecular mass = 28 + 80 = 108
108 g of N₂O₅ contains 6.023 x 10²³ molecules.
∴ 54 g of N₂O₅ will contain \(\frac{6.023 \times 10^{23}}{108}\) x 54 = 3.0115 x 10²³ molecules.

(b) 28 g of CO₂
CO₂ = Molecular mass = 12 + 32 = 44
44 g of CO₂ contains 6.023 x 10²³ molecules.
∴ 28 g of CO₂ will contain \(\frac{6.023 \times 10^{23}}{44}\) x 28 = 3.832 x 10²³ molecules.

(c) 36 g of H₂O
H₂O = Molecular mass = 2 + 16 = 18
18 g of H₂O contains 6.023 x 10²³ molecules.
∴ 36 g of H₂O will contain \(\frac{6.023 \times 10^{23}}{18}\) x 36 = 12.046 X 10²³molecules.

(d) 46 g of C₂H₅OH
C₂H₅OH = Molecular mass = 24 + 6 + 16 = 46
46 g of C₂H₅OH contains 6.023 x 10²³ molecules.
So, among the 4, 36 g of water contain the largest number of molecules as 12.046 x 10²³.

Question 4.
The number of moles of H₂ in 0.224 liter of hydrogen gas at STP is …………..
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(c) 0.01
22.4 liter of hydrogen gas at STP contains 1 mole.
∴ 0.224 liter of hydrogen gas at STP will contain \(\frac{1}{22.4}\) x 0.224 = 0.01

Question 5.
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be ………..
(a) 3 mol
(b) 4 mol
(c) 1 mol
(d) 2 mol
Answer:
(b) 4 mol
10 g of H₂ + 64 g of O₂ → water
2H₂ + O₂ → 2H₂O
4 g of hydrogen react with 32 g of oxygen. So 10 g of hydrogen will react with 80 g of oxygen. But we have given amount of oxygen only 64 g. It means, here oxygen is the limiting agent. Now all the oxygen react with 8 g of hydrogen and form 4 moles of water.

Question 6.
6.023 x 10²⁰ molecules of urea are present in 100 ml of its solution. The concentration of the solution is –
(a) 0.02 M
(b) 0.1 M
(c) 0.01 M
(d) 0.001 M
Answer:
(c) 0.01 M
100 ml of solution contain 6.023 x 10²⁰ molecules.
6.023 x 10²³ molecules in 1000 ml = 1 M.
∴ 6.023 x 10²⁰ molecules in 1000 ml = 
= 10²⁰ x 10^-23 = 10^-3
10^-3 moles present in 100 ml.
∴ In 1000 ml the moles present is = \(\frac{10^{-3}}{100}\) x 1000 = 10^-2 moles
Concentration = 0.01 M

Question 7.
Two containers A and 13 of equal volume contain 6 g of O₂ and SO₂ at 300 K and 1 atm. Then
(a) No. of molecules in A is less than that in B
(b) No. of molecules in A is more than that in B
(c) No. of molecules in A and B are same
(d) none of these
Answer:
(b) No, of molecules in A is more than that in B
O₂ = Mass = 6 g
Molar mass = 32 g
32 g of O₂ contains 6.023 x 1023 molecules.
∴ 6 g of O₂ will contain = 6.023 X 1023 x 6 = 1.129 x 10²³ molecules.
SO₂ = Mass = 6 g
Molar mass = 64 g
64 g of SO₂ contains 6.023 x 10²³ molecules.
∴ 6 g of SO₂ will contain 6.023 x 1023 x 6 = 0.5646 x 1023 molecules.
∴ Number of molecules in A is more than that in B.

Question 8.
The number of molecules in 16 g of methane is ……….
(a) 3.023 x 10²³
(b) 6.023 x 10²³
(c) 16 / 6.023 x 10²³
(d) 6.023 / 3 x 10²³
Answer:
(b) 6.023 x 10²³
Hint:
Methane: CH₄
Molecular mass 12 + 4 = 16
16 g of methane contains Avogadro number of molecules 6.023 x 10²³ molecules.

Question 9.
Number of atoms in 4.25 g of ammonia is …………..
(a) 1 x 10²³
(b) 2 x 10²³
(e) 4 x 10²³
(d) 6 x 10²³
Answer:
(d) 6 x 10²³
Hint:
Ammonia = NH₃ (4 atoms)
Molecular mass = 14 + 3 = 17
17 g of Ammonia contains 6.023 X 10²³ atoms.
∴ 4.25 g of Ammonia will contain 6.0231; 1023 x 4.25 = 1.5055 x 10²³ molecules.
∴ 1.5055 x 10²³ molecules will contain 4 x 1.5 x 10²³ = 6 x 10²³ molecules.

Question 10.
The number of molecules in a drop of water (0.0018 ml) at room temperature is ……….
(a) 6.02 x 10²³
(b) 1.084 x 10²³
(c) 4.84 x 10²³
(d) 6.02 x 10²³
Answer:
(a) 6.02 x 10²³
Hint:
0.0018 ml – drop of water = 0.0018 g
H₂O = molecular mass = 18 g.
Number of molecules in 18 g = 6.023 x 10²³
∴ Number of molecules in 0.0018 g 6.023 x 1023 x 0.0018 = 6.023 x 10²³ x 10⁵
= 6.023 x l0¹⁹ molecules.
or
Density of water at 25°C = 997.0479 g / L
Mass of 0.0018 ml (or) 0.00 18 x 10^-3 L
= D x V
= 997.05 x 0.0018 x 10^-3
= 1.795 x 10^-3 g
Molar mass of water = 18 g
Mole = \(\frac{Mass}{Molecular mass}\) = \(\frac{1.795 \times 10^{-3}}{18}\)
= 9.971 x 10^-5
∴ Number of molecules in 0.0018 ml = moles x Avogadro number
= 9.971 x 10^-5 x 6.023 x 10²³
= 6 x 10¹⁹ molecules.

Question 11.
7.5 g of a gas occupy 5.6 liters of volume at STP. The gas ……….
(a) NO
(b) N₂O
(c) CO
(d) CO₂
Answer:
(a) NO
Hint:
22.4 liters = 1 mole
5.6 liters = \(\frac {1}{22.4}\) x 5.6 = 0.25 mole.
NO = molar mass = 14 + 16 = 30 g = 1 mole
N₂O = molar mass = 28 + 16 = 44g = 1 mole
CO = molar mass = 12 + 16 = 28 g = 1 mole
CO₂=molar mass = 12 + 32 = 44g = 1 mole
Among the four gases, 0.25 mole = 7.5 g is equal to NO gas.

Question 12.
The mass in grams of 0.45 mole of CO₂ ions ……….
(a) 1.8
(b) 40
(c) 36
(d) 18
Answer:
(d) 18
Hint:
Ca = Atomic mass = 40
Ca → Ca²⁺ + 2e^–
41 g of Ca = 1 mole (for Ca²⁺ Atomic mass remains same)
1 mole of Ca²⁺ = 40 g
∴ 0.45 mole of Ca²⁺ = \(\frac {40}{1}\) x 0.45 = 18g

Question 13.
The mass of one molecule of HI in grams is ……….
(a) 2.125 x 10^-22
(b) 128
(c) 127
(d) 6.02 x 10^-23
Answer:
(b) 128
Hint:
HI = 1 mole = 1 + 127 = 128 g

Question 14.
Avogadro’s number is the number of molecules present in ……….
(a) 1 g of molecule
(b) 1 g atom of molecule
(c) gram molecular mass
(d) I lit of molecule
Answer:
(c) gram molecular mass
by definition (c) is correct

Question 15.
Which of the following contains same number of carbon atoms as are in 6.0 g of carbon (C-12)?
(a) 6.0 g ethane
(b) 8.0 g methane
(c) 21.0 g Propane
(d) 28.0 g CO
Answer:
(b) 8.0 g methane

Hint:
(a) 6.0 g of ethane (C₂H₆)
C₂H₆ = molar mass = 24 + 6 = 30 g
30 g of ethane contains 2 x 6.023 x 10²³ Carbon atoms.

(b) 8.0 g of methane (CH₄)
CH₄ molar mass = 12 + 4 = 16g
16 g of methane contains 6.023 x 10²³ Carbon atoms.

(c) 21.0 g of propane (C₃H₈)
C₃H₈ = molar mass = 36 + 8 = 44 g
44 g of propane contains 3 x 6.023 x 10²³ Carbon atoms.

(d) 28.0 g of Carbon monoxide (CO)
CO = molar mass = 12+ 16 = 28 g
28 g of Carbon monoxide contains 6.023 x 10²³ Carbon atoms.
6.0 g of Carbon contains = 6.023 x 10 x 6 = 3.0115 x 10²³ Carbon atoms.
Among the (a), (b), (c), (d) – 8 g of CH₄ contains x 8 = 3.0115 x 10²³ Carbon atoms.

Question 16.
Equivalent mass of KMnO₄ when it is converted to MnSO₄ is equal to molar mass divided by ………..
(a) 6
(b) 4
(c) 5
(d) 2
Answer:
(c) 5
Hint:

Question 17.
How many equivalents of Sodium sulphate is formed when Sulphuric acid is completelyn neutralized with a base NaOH?
(a) 0.2
(b) 2
(e) 0.1
(d) 1
Answer:
(d) 1
Hint:

Question 18.
Cl₂ changes to Cl^– and ClO^– in cold NaOH. Equivalent mass of Cl₂ will be ………..
(a) Molar mass / 2
(b) Molar mass / 1
(c) Molar mass / 3
(d) 2 x Molar mass / 2
Answer:
(a) Molar mass / 2

Question 19.
Equivalent mass of KMnO₄ in acidic medium, concentrated alkaline medium and dilute basic medium respectively are M, M, M. Reduced products can be ……………
(a) MnO₂, MnO₂^2-, Mn²⁺
(b) MnO₂, Mn²⁺, MnO₄^2-
(c) Mn²⁺, MnO₂, MnO₄^2-
(d) Mn²⁺, MnO₄^2-, MnO₂
Answer:
(c) Mn²⁺, MnO₂, MnO₄^2-
Hint:
MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O (acidic medium)
MnO₄ + 4H⁺ + 3e^– → MnO₂ + 2H₂O (concentrated basic medium)
MnO₄ + e^– → MnO₄^2- (dilute basic medium)

Question 20.
The empirical formula of hydrogen peroxide is
(a) HO
(b) H₂O
(e) H₃O
(d) H₂O₂
Answer:
(a) HO
Hint:
Molecular formula of hydrogen peroxide = H₂O₂
H₂O₂ ÷ 2 = HO = Empirical formula.

Question 21.
Molecular mass =
(a) Vapour Density × 2
(b) Vapour Density ÷ 2
(c) Vapour Density × 3
(d) Vapour Density
Answer:
(a) Vapour Density × 2

Question 22.
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage of purity of magnesium carbonate in the sample?
(a) 60
(b) 84
(e) 75
(d) 96
Answer:
(b) 84
Hint:

84 g of MgCO₃ gives 40 g of MgO. (100% purity)
20 g of MgCO₃ will give = \(\frac{40 \times 20}{84}\) = 9.52 g
9.52 g of MgO is given by 100% pure MgCO₃.
8.0 g of MgO will be given by = \(\frac{100 \times 8}{9.52}\) = 84.03%

Question 23.
What is the mass of the precipitate formed when the preparation of alkyl halides 50 ml of 16.9 % solution of AgNO₃ is mixed with 50 ml of 5.8 % NaCl solution?
(a) 7 g
(b) 14 g
(c) 28 g
(d) 35 g
Answer:
(a) 7 g
Hint:

Question 24.
When 22 L of hydrogen gas is mixed with 11.2 L of chlorine gas, each at STP, the moles of HCl gas formed is equal to ……….
(a) 2
(b) 0.5
(c) 1.5
(d) 1
Answer:
(d) 1
Hint:
H₂ + Cl₂ → 2HCl
22 liters + 11.2 litres
1 mole + \(\frac {1}{2}\) mole = 1 mole of HCl and \(\frac {1}{2}\) mole of H₂ is remained.

Question 25.
5.6 L of a gas at STP are found to have mass of 11 g. The molecular mass of the gas is
(a) 36
(b) 48
(c) 40
(d) 44
Answer:
(d)44
Hint:
5.6 L of gas has the mass of 11 g.
∴ 22.4 L of gas will have the mass \(\frac {11}{5.6}\) x 22.4 = 44

Question 26.
Oxidation number f Fluorine in all compounds is ……….
(a) + 1
(b) -1
(c) 0
(d) – 2
Answer:
(b) – 1

Question 27.
In redox reaction which of the following is true?
(a) Number of electrons lost is more than number of electrons gained
(b) Number of electrons lost is less than number of electrons gained
(c) Number of electrons lost s equal number of electrons gained
(d) No transfer and gain of electrons during the reaction.
Answer:
(c) Number of electrons lost is equal number of electrons gained

Question 28.
Which of the following is a mono-atomic molecule?
(a) Hydrogen
(b) Oxygen
(c) Sodium
(d) Ozone
Answer:
(c) Sodium

Question 29.
Which one of the following is a diatomic molecule?
(a) Ozone
(b) Copper
(c) Hydrogen
(d) Gold
Answer:
(c) Hydrogen

Question 30.
The value of Avogadro Number N is equal to ……….
(a) 2.24 x 10^-2L
(b) 22400 cm³
(c) 6.023 x 10^-23
(d) 6.023 x 10²³
Answer:
(d) 6.023 x 10²³

Question 31.
46 g of ethanol contains ……….
(a) 2 x 6.023 x 10²³ C atoms
(b) 3 x 6.023 x 10²³ atoms
(c) 9 x 6.023 x 10²³ H atoms
(d) 6.023 x 10²³ Carbon atoms
Answer:
(a) 2 x 6.023 x 10²³ C atoms
Hint:
C₂H₅OH = Molecular mass = (12 x 2) + (1 x 6) + (1 x 16)
=24 + 6 + 16 = 46
2 Carbon atoms are present.
∴ 2 x 6.023 x 10²³ C atoms is correct.

Question 32.
The mass of one mole of CaCl₂ is ……….
(a) 55.5 g mol^-1
(b) 111 g mol^-1
(c) 222 g mol ^-1
(d) 77.5 g mol^-1
Answer:
(b) 111 g mol^-1
Hint:
CaCl₂ = Mass = 40 + 71 = 111 g mol^-1

Question 33.
22 g of CO₂ contains molecules of CO₂
(a) 6.023 x 10²³
(b) 6.023 x 10²³
(c) 3.0115 x 10²³
(d) 3.0115 x 10²³
Answer:
(c) 3.0115 x 10²³
Hint:
44 g of CO₂ contains 6.023 x 10²³ molecules.
∴ 22 g of CO₂ will contain = \(\frac{6.023 \times 10^{23}}{44}\) x 22 = 3.0115 x 10²³

Question 34.
The formula weight of ethanol (C₂H₅OH) is ……….
(a) 56.5 amu
(b) 16 amu
(c) 60 amu
(d) 46 amu
Answer:
(d) 46 amu
Hint:
Formula weight of C₂H₅OH = (12 x 2) + (6 x 1) + (1 x 16)
= 24 + 6 + 16 = 46 amu

Question 35.
The number of moles of ethane in 60 g is ……….
(a) 2
(b) 4
(c) 0.5
(d) 1
Answer:
(a) 2
Hint:
C₂H₆ – Ethane – molar mass = 24 + 6 = 30 g
30 g of C₂H₆ contains = 1 mole.
∴ 60 g of C₂H₆ will contains = x 1 = 2 moles.

Question 36.
Which of the following method is used to prevent rusting of iron?
(a) Galvanization
(b) Painting
(c) Chrome plating
(d) all the above
Answer:
(d) all the above

Question 37.
Which of the following is not a redox reaction?
(a) H₂ + F₂ → 2HF
(b) Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
(c) 2H₂ + O₂ → 2H₂O
(d) AgCl + NH₃ → [Ag(NH₃)₂]Cl
Answer:
(d) AgCl + NH₃ → [Ag(NH₃)₂]Cl

Question 38.
How many H₂O molecules are there in a snowflake weighing 1 mg?
(a) 3.35 x 10¹⁹
(b) 6.023 x 10²³
(c) 335 x 10^-19
(d) 100
Answer:
(a) 3.35 x 10¹⁹
Hint:
1 mg of H₂O
Molar mass of H₂O = 2 + 16 = 18 g
1 Mole contains 6.023 x 10²³ water molecules
18 g contains 6.023 x 10²³ water molecules
l mg contains \(\frac{6.023 \times 10^{23}}{18}\) x 1g /1000 mg = 3.35 x 10^-19 H₂O molecule.

Question 39.
The volume of HCl gas weighing 73 g at STP is ……….
(a) 2.24 x 10^-2 m³
(b) 4.48 x 10^-2 m³
(c) 4.48 x 10²m³
(d) 2.24 x 10²m³
Answer:
(b) 4.48 x 10^-2 m³
Hint:
HCl = Molar mass = 1 + 35.5 = 36.5 g
36.5 g of HCl occupies 2.24 x 10^-2m³
∴ 73 g of HCl at STP will occupy \(\frac{2.24 \times 10^{-2}}{36.5}\) x 73 = 4.48 x 10^-2m³

Question 40.
The molar volume of 22 g of CO₂ is ……….
(a)2.24 x 10^-2m³
(b)4.48 x 10^-2m³
(c) 1.12 x 10^-2m³
(d)2.24 x 10^-2m³
Answer:
(c) 1.12 x 10^-2m³
Hint:
CO₂ = Molar’mass of 12 + 32 = 44 g
44 g of CO₂ occupies molar volume = 2.24 x 10^-2m³
∴ 2g of CO₂ will occupy = \(\frac{2.24 \times 10^{-2}}{44}\) x 2= 1.12 x 10^-2 L

Question 41.
The equivalent mass of Aluminium is ……….
(a) 27
(b) 13.5
(c) 54
(d) 9
Answer:
(d) 9

Question 42.
The equivalent mass of HSO₄ is ……….
(a) 98
(b) 97
(c) 48
(d) 96
Answer:
(a) 98
Hint:
HSO₄ = Molar mass= 1 + 32 + 64 + 1 = 98
Equivalent mass = Molar mass / 1 = 98

Question 43.
The equivalent mass of NaCl is ……….
(a) 40
(b) 58.5
(c) 35.5
(d) 23
Answer:
(b) 58.5
Hint:
NaCl = Salt Molar mass 23 + 35.5 = 58.5
Equivalent mass of Salt = Molar mass of Salt.

Question 44.
Match the List-I and List-lI using the correct code given below the list.

Answer:

Question 45.
Match the List-I and List-Il using the correct code given below the list.


Answer:

Question 46.
Consider the following statements ……….
(i) Empirical formula shows the actual number of atoms of different elements in one molecule of the compound.
(ii) Ozone is a diatomic molecule.
(iii) Gases are easily compressible.
Which of the above statement is/are not correct?
(a) (i), (ii), (iii)
(b) (i) & (ii)
(c) (ii) & (iii)
(d) (iii) only
Answer:
(b) (i) & (ii)

Question 47.
How many molecules of hydrogen is required to produce 4 moles of ammonia?
(a) 15 moles
(b) 20 moles
(c) 6 moles
(d) 4 moles
Answer:
(c) 6 moles
Hint:
3H₂ + N₂ → 2NH₃
To get 2 moles of ammonia, 3 mole of H₂ is required.
To get 4 moles of ammonia = \(\frac {3}{2}\), x 4
= 6 moles of H₂ is required.

Question 48.
The number of moles of oxygen required to prepare 1 mole of water is …………..
(a) I mole
(b) 0.5 mole
(c) 2 moles
(d) 0.4 mole
Answer:
(b) 0.5 mole
Hint:

0.5 mole of oxygen is required to prepare 1 mole of H₂O.

Question 49.
How much volume of CO₂ is produced when 50 g of CaCO₃ is heated strongly?
(a) 2.24 x 10^-2 m³
(b) 22.4
(c) 11.2 L
(d) 22400 cm³
Answer:
(c) 11.2 L

100 g of CaCO₃ produces 22.4 litres of CO₂.
50 g of CaCO₃ will produce = \(\frac{22.4}{100} \times 50\) = 11.2 litres

Question 50.
Which one of the following is not a redox reaction?
(a) Rusting of iron
(b) Extraction of metal Na
(c) Electroplating
(d) Aluminothermic process
Answer:
(a) Rusting of Iron

Question 51.
In the reaction 2 AuCl₃ + 3 SnCl₂ → 2 Au + 3 SnCl₄ which is an oxidising agent?
(a) AuCl₃
(b) Au
(c) SnCl₂
(d) Both AuCl₃ and SnCl₂
Answer:
(a) AuCl₃
Hint:
AuCl₃ undergoes reduction. So, it is an oxidising agent.

Question 52.
Identify the compound formed during the rusting of iron.
(a) Fe₂O₃
(b) Fe₂O₃. x H₂O
(c) FeO. x H₂O
(d) FeO
Answer:
(b) Fe₂O₃. x H₂O – Hydrated iron oxide is rust.

Question 53.
The oxidation state of a substance in its elementary state is equal to ………..
(a) -1
(b) -2
(c) zero
(d) charge of the ion
Answer:
(c) zero

Question 54.
The oxidation number of fluorine in all its compounds is equal to ………….
(a) -1
(b) +1
(c) – 2
(d) +2
Answer:
(a) -1

Question 55.
Consider the following statements.
(i) The sum of the oxidation number of all the atoms in neutral molecule is equal to zero.
(ii) Fluorine has an oxidation number +1 in all its compounds. .
(iii) The oxidation number of a substance in its elementary state is equal to zero.
Which of the above statement is/are not correct?
(a) (i), (ii) & (iii)
(b) (ii) & (iii)
(c) (i) only
(d) (ii) only
Answer:
(d) (ii) only

Question 56.
The oxidation number of Cr in K₂Cr₂O₇ is ………….
(a) +4
(b) + 6
(c)  O
(d) + 7
Answer:
(b) + 6
K₂Cr₂O₂
2 + 2x – 140
2x – 12 = 0
2x = + 12
x = + 6

Question 57.
The oxidation number of N in NH₄ ion is ……….
(a) +4
(b) + 3
(c) – 3
(d) – 4
Answer:
(c) – 3
NH₂⁺
x + 4 = + 1
x = + 1 – 4
x = – 3

Question 58.
Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s). In this reaction, which gets oxidised?
(a) Cu²⁺
(b) Zn²⁺
(c) Zn
(d) Zn, Cu²⁺
Answer:
(c) Zn
Hint:
Zn → Zn²⁺ + 2e^– (loss of electron = oxidation)
Zn gets oxidised

Question 59.
Which one of the following is an example for disproportionation reaction?
(a) CuSO₄ + Zn → ZnSO₄ + Cu
(b) 2KClO₃ → 2KCI + 3O₂
(c) PCl₅ → PCl₃ + Cl₂
(d) 4H₃PO₃ → 3H₃PO₄ + PH₃
Answer:
(d) 4H₃PO₃ → 3H₃PO₄ + PH₃ (Auto oxidation and reduction reaction)

Question 60.
The number of molecules in 40 g of sodium hydroxide is ……….
(a) 6.023 x 10²³
(b) 3.0115 x 10²³
(c) 6.023 x 10²³
(d) 2 x 6.023 x 10²³
Answer:
(c) 6.023 x 10²³
Sodium hydroxide = NaOH = 23 + 16+ 1 =40 g
40g = 1 mole = 6.023 10²³

Question 61.
The mass of one molecule of AgCl in grams is ……….
(a) 108 g
(b) 143.5 g
(c) 35.5 g
(d) 243.5 g
Answer:
(b) 143.5 g
Hint:
Mass of AgCl = 108 + 35.5 = 143.5 g.

Question 62.
The empirical formula of Alkene is ……….
(a) CH
(b) CH₂
(c) CH₃
(d) CH₃O
Answer:
(b) CH₂
Hint:
Alkene C_nH_2n Molecular formula
E.F. = M.F./2
∴ Empirical formula = CH₂

Question 63.
22 g of a gas occupies 11.2 litres of volume at STP. The gas is ……….
(a) CH₄
(b) NO
(c) CO
(d) CO₂
Answer:
(d) CO₂
Hint:
22 g of a gas occupies 11.2 litres.
11.2 liters is occupied by 22 g of a gas.
∴ Molar volume 22.4 liter will be occupied by \(\frac {22}{11.2}\) x 22.4 = 44 g
∴ The gas is CO₂.

Question 64.
The number of moles of H₂ in 2.24 liter of hydrogen gas at STP is ……….
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(b) 0.1
Hint:
22.4 liter = 1 molar volume = 1 mole.
∴ 2.24 liter = \(\frac {1}{22.4}\) x 2.24 = 01 mole

Question 65.
How many molecules are present in 32 g of methane?
(a) 2 x 6.023 x 10²³
(b) 6.023 x 10²³
(c) 6.023 x 10²³
(d) 3.011 x 10²³
Answer:
(a) 2 x 6.023 x 10²³
Hint:
Methane (CH₄) – Molar mass = 12 + 4 = 16g.
16 g contains 6.023 x 10²³ molecules.
∴ 32 g of methane will contain = \(\frac{6.023 \times 10^{23}}{16} \times 32^{2}\) = 2 x 6.023 x 10²³

Question 66.
The empirical formula of glucose is ……….
(a) CH
(b) CH₂O
(c) CH₂O₂
(d) CHO
Answer:
(b) CH₂O
Hint:
Glucose = Molecular formula = C₆H₁₂O₆
Empirical formula = \(\frac { Molecular formula}{6}\) = \(\\frac{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{6}\) = CH₂O

Question 67.
How many moles of water is present in I L of water?
(a) 1
(b) 18
(c) 55.55
(d) 5.555
Answer:
(c) 55.55
Hint:
1 Liter of water = 1000 g.
Molar mass of water 18 g
Number of moles = \(\frac {Mass}{Molar mass}\) = \(\frac {1000}{18}\) = 55.55 moles

Question 68.
How many moles of Hydrogen atoms are present in 1 mole of C₂ H₆ ?
(a) 18 moles
(b) 6 moles
(c) 3 moles
(d) 1 mole
Answer:
(b) 6 moles
Hint:
C₂H₆ contains 6H atoms. 6 moles.

Question 69.
The molar mass of Na₂SO4 is ……….
(a).129
(b) 142
(c) 110
(d) 70
Answer:
(b) 142
Hint:
Na₂ SO₄ = Molar mass
= (23 x 2) + (32 x 1) + (16 x 4)
= 46 + 32 + 64 = 142

Question 70.
Match the List-I with List-Il using the correct code given below the list.

Answer:

Question 71.
Ore mole of CO₂ contains ………….
(a) 6.02 x 10²³ atoms of C
(b) 3 g of CO₂
(c) 6.02 10²³ atoms of O
(d) 18.1 x 10²³ molecules of CO₂
Answer:
(a) 6.02 x 10²³ atoms of C

Question 72.
5.6 liters of oxygen at STP is equivalent to ……….
(a) 1 mole
(b) 1/4 mole
(c) 1/8 mole
(d) 1/2 mole
Answer:
(b) 1/4 mole
Hint:
22.4 litres of O₂ = 1 mole
∴ 5.6 litres of O₂ = \(\frac {1}{22.4}\) x 5.6 = 0.25 mole = 1/4 mole.

Question 73.
How many grams are contained in 1 gram atom of Na?
(a) 13 g
(b) 1 g
(c) 23 g
(d) 1/23 g
Answer:
(c) 23 g
Hint:
1 gram atom of Na
Na = Atomic mass 23 g (or) 23 amu
1 gram atom of Na = 1 mole = 23 g.

Question 74.
12 g of Mg will react completely with an acid to give ……….
(a) 1 mole of O₂
(b) 1/2 mole of H₂
(c) I mole of H₂
(d) 2 mole of H₂
Answer:
(b) 1/2 mole of H₂
Hint:

∴ 12 g of Mg \(\frac {1}{2}\) mole of H₂O

Question 75.
which of the following has the highest mass?
(a) I g atom of C
(b) 1/2 mole of CH₄
(c) 10 ml of water
(d) 3.011 x 10²³ atoms of oxygen
Answer:
(a) I g atom of C
(a) 1 g atom of C = 12 g
(b) 1/2 mole of CH₄ = \(\frac {12 + 4}{2}\) = 8 g.
(c) 10 ml of water (H₂O) = 1 x 10 = 10 g
(d) 3.011 x 10²³ atoms of oxygen = 0.5 mole of oxygen = 8 g

Question 76.
The empirical formula of sucrose is ……….
(a) CH₂O
(b) CHO
(c) C₁₂H₂₂O₁₁
(d) C(H₂O)₂
Answer:
(a) CH₂O
Hint:
Sucrose Molecular formula = C₁₂H₂₂O₁₁
E.F.= \(\frac{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}{12}\) = CH₂O

Question 77.
The number of grams of oxygen in 0.10 mol of Na₂CO₃. 10H₂O is ………..
(a) 20.8 g
(b) 18 g
(c) 108 g
(d) 13 g
Answer:
(a) 20.8 g
Hint:
Na₂CO₂.10H₂O = 1 mole
1 mole of Na₂CO₃. 10H₂O contains 13 oxygen atoms.
Mass of 13 oxygen atoms = 13 x 16 = 208
1 mole of Na₂CO₃.10H₂O contains 208 g of oxygen.
∴ 0.10 mole of Na₂CO₃.10H₂O contains \(\frac {208}{1}\) x 0.12 = 08 g.

Question 78.
The mass of an atom of nitrogen is ……….

Answer:

Question 79.
Which of the following halogens do not exhibit positive oxidation number in its compounds?
(a) Fluorine
(b) Chlorine
(c) Iodine
(d) Bromine
Answer:
(a) Fluorine

Question 80.
Which of the following is the most powerful oxidising agent?
(a) KMnO₄
(b) K₂Cr₂O₇
(c) O₃
(d) H₂O₂
Answer:
(a) KMnO₄

Question 81.
On the reaction 2Ag + H₂SO₄ → Ag₂SO₄ + 2H₂O + SO₂. Sulphuric acid acts as ……………
(a) oxidising agent
(b) reducing agent
(c) a catalyst
(d) an acid as well as an oxidant
Answer:
(d) an acid as well as an oxidant

Question 82.
The oxidation number of carboxylic carbon atom in CH₃COOH is ……….
(a) + 2
(b) + 4
(c) + 1
(d) + 3
Answer:
(d) + 3
CH₃COOH
-3 + 3 + x – 4 + 1
x – 3 = 0
x = + 3
Carboxylic carbon oxidation number = + 3

Question 83.
When methane is burnt in oxygen to produce CO₂ and H₂O, the oxidation number of carbon changes by ……….
(a) – 8
(b) + 4
(c) zero
(d) + 8
Answer:
(b) + 4

Question 84.
The oxidation number of carbon is zero in ……….
(a) HCHO
(b) C₁₂H₂₂O₁₁
(c) C₆H₁₂O₆
(d) all the above
Answer:
(d) all the above

Question 85.
The oxidation number of Fe in Fe₂(SO₄)₃ is ……….
(a) + 2
(b) + 3
(c) + 2, + 3
(d) O
Answer:
(b) + 3
Fe₂(SO₄)₃
2x – 6 = 0
x = + 3

Question 86.
Among the following molecules in which Chlorine shows maximum oxidation state?
(a) Cl₂₁
(b) KCl
(c) KClO₃
(d) Cl₂O₇
Answer:
(d) Cl₂O₇
Cl₂O₇
2x – 14 = 0
2x = + 14
x = + 7

Question 87.
The oxidation number of carbon in CH₃ → CH₂OH is ………….
(a) + 2
(b) – 2
(e) O
(d) + 4
Answer:
(b) – 2
C₂H₅OH
2x +6 – 2 = O
2x + 4 = 0
2x = -4
x = -2

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 2 – Mark Questions

I. Write brief answer to the following questions:

Question 1.
State Avogadro’s Hypothesis.
Answer:
It states that ‘Equal volume of all gases under the same conditions of temperature and pressure contain the same number of molecules’.

Question 2.
What is molar volume?
Answer:
Molar volume is the volume occupied by one mole of a substance in the gaseous state at STP. It is equal to 2.24 x 10^-2m³ (22.4 L).

Question 3.
The approximate production of Na₂CO₃ per month is 424 x 10⁶ g while that of methyl alcohol is 320 x 10⁶ g. Which is produced more in terms of moles?
Answer:
Na₂CO₃ mass = 424 x 10⁶g
Molecular mass of Na₂CO₃ = (23 x 2) + 12 + (16 x 3)
= 46+ 12 +48
= 106 g
No. of moles of Na₂CO₃
= 4 x 10⁶ moles
Methyl alcohol mass = 320 x 10⁶ g
Molecular mass of CH₃OH = 12 + (1 x 4)+ 16 = 32 g
= 12 + 4 + 16 = 32 g
No. of moles of Methyl alcohol
= 10 x 10⁶ moles.
∴ Methyl alcohol is more produced in terms of moles.

Question 4.
Calculate number of moles of carbon atoms ¡n three moles of ethane.
Answer:
Ethane – Molecular formula = C₂H₆
1 mole of ethane contains 2 atoms of carbon (6.023 x 10²³ C)
∴ 3 moles of ethane contains 6 atoms of Carbon.
∴ No. of moles of Carbon atoms = 3 x 6.023 x 10²³ Carbon atoms.
= 18.069 x 10²³ Carbon atoms.

Question 5.
Find the molecular mass of FeSO₄.7H₂O
Answer:
Sum of Atomic mass of all elements = Molecular mass
Atomic mass of Fe = 56.0
Atomic mass of S = 32.0
Atomic mass of 4[O] = 64.0
Atomic mass of 14[H] = 14.0
Atomic mass of 7[O] = 112.0 = 278.0
Molecular mass of FeSO₄.7H₂O = 278 g.

Question 6.
Mass of one atom of an element ¡s 6.66 x 10²³ g. How many moles of element are there in 0.320 kg?
Answer:
Mass of one atom of an element = 6.66 x 10²³g
No. of moles = \(\frac {Mass}{Molecular mass}\) 3
Molecular mass = Mass of 1 atom x Avogadro number
6.66 x 10²³ x 6.023 x 10²³
= 6.66 x 6.023 = 40.11318
Number of moles = \(\frac {Mass}{Molecular mass}\) = \(\frac{0.320 \mathrm{kg} \times 10^{3}}{40}\) = 8 moles.

Question 7.
How many moles of glucose are present in 720 g of glucose?
Answer:
Glucose = C₆H₄O₄
Molecular mass of Glucose = (12 x 6) + (1 x 12) + (16 x 6)
= 72 + 12 + 96 = 180
\(\frac {720}{180}\) = 4 moles.

Question 8.
Calculate the weight of 0.2 mole of sodium carbonate.
Answer:
Sodium carbonate = Na₂CO₃
Molecular mass of Na₂CO₃ = (23 x 2)+(12 x 1)+(16 x 3)
= 46 + 12 + 48 = 106 g
Mass of 1 mole of Na₂CO₃ = \(\frac{106 \times 0.2}{1}\) = 21.2 g

Question 9.
What do you understand by the terms acidity and basicity?
Answer:
Acidity:
The number of hydroxyl ions present in one mole of a base is known as the acidity of the base.

Basicity:
The number of replaceable hydrogen atoms present in a molecule of the acid is referred to as its basicity.

Question 10.
Calculate the equivalent mass of bicarbonate ion.
Answer:
Bicarbonate ion = HCO₃

Formula mass of HCO₃ = 1 + 12 + 48 = 61
Equivalent mass of HCO₃ = \(\frac {61}{1}\) = 61

Question 11.
Calculate the equivalent mass of barium hydroxide.
Barium hydroxide = Ba(OH)₂
Molecular mass of Ba(OH)₂ = 137 + (16 x 2) + (1 x 2)
= 171.0 g / mol.
Acidity = 2
Equivalent mass of Ba(OH)₂ =
= \(\frac {17 1.0}{2}\) = 85.5

Question 12.
Calculate the equivalent mass of hydrated sodium carbonate.
Answer:
Hydrated sodium carbonate = Na₂CO₃. 10H₂O
Molecular mass of Na₂CO₃. 10H₂O = (23 x 2) + (2 x 1) + (16 x 13) + (1 x 20)
= 46 + 12 + 208 + 20 = 286
Equivalent mass of Na₂CO₃. 10H₂O = \(\frac {Molecular mass}{Acidity}\)
= \(\frac {286}{2}\) = 143

Question 13.
What do you understand by the terms empirical formula and molecular formula?
Answer:
Empirical Formula:

• It is the simplest formula.
• It shows the ratio of number of atoms of different elements in one molecule of the compound.

Molecular Formula:

• It is the actual formula.
• It shows the actual number of different types of atoms present in one molecule of the compound.

Question 14.
Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?
Answer:
Molecular mass of H₃BO₃ = (1 x 3) + (11 x 1) + (16 x 3) = 62
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass x mole
= 62 x 0.543
= 33.66 g

Question 15.
A compound contains 50% of X (atomic mass 10) and 50% Y (atomic mass 20). Give its molecular formula
Answer:

∴ The Empirical Formula is X₃Y
Empirical Formula mass = 20 + 20 = 40
Molecular mass = Sum of atomic mass = 40
n = 1, Molecular formula = (Empirical Formula )_n = (X₂Y)₁ = X₂Y.

Question 16.
Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
Answer:
Sodium Nitrate = NaNO₃
Molecular mass of Sodium Nitrate = 23 + 14 + 48 = 85
100% pure 85 g of NaNO₃ contains 23 g of Sodium.
100% pure 95 x 103 g of NaNO₃ will contains \(\frac {23}{85}\) x 95 x 10³
= 25.70 x 10³ g of Sodium.
100% pure NaNO₃  contains 25.70 x 10³ g of Sodium.
∴ 70% pure NaNO₃  will contains = img
= 17990 g (or) 17.99 Kg of Na.

Question 17.
Define matter. What are the types of matter?
Answer:

• A matter is anything which has mass and occupies space.
• Matters exist in all three states such as solid, liquid and gas.

Question 18.
Prove that states of matter are inter convertible.
Answer:
States of matter are inter convertible by changing temperature and pressure.

Question 19.
What is meant by Plasma state? Give an example.
Answer:
Gaseous state of matter at very high temperature containing gaseous ions and free electron is referred to as the Plasma state. e.g. Lightning.

Question 20.
Differentiate an element and an atom.
Answer:

• An atom is the ultimate smallest electrically neutral, being made up of fundamental particles such as proton, neutron and electron.
• An element consists of only one type of atoms. Elements are further divided into metals, non-metals, and metaloids.

Question 21.
Distinguish between a molecule and a compound
Answer:
Molecule:

• A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence.
• e.g. Na – Mono atomic molecule O₂ – Diatomic molecule P₄ – Poly atomic molecule

Compound:

• A molecule which contains two or more atoms of different elements are called a compound molecule.
• e.g. CO₂ – Carbon dioxide CH₄ – Methane H₂O – Water

Question 22.
Chlorine has fractional average atomic mass. Justify this statement.
Answer:
Chlorine molecule has two isotopes as in ₁₇Cl³⁵ , ₁₇ Cl³⁷ in the ratio of 77 : 23, so when we are calculating the average atomic mass, it becomes fractional.
The average relative atomic mass of Chlorine = \(\frac {(35 x 77) + (37 x 23)}{100}\) = 35.46 amu

Question 23.
Define molecular mass of a substance.
Answer:
Molecular mass of a substance (element or compound) represents the number of times the molecule of that substance is heavier than 1 / 12^th of the mass of an atom of C-12 isotope. Molecular mass = 2 x Vapour density

Question 24.
Calculate the molecular mass of Sulphuric acid (H₂SO₄). Element
Answer:

Question 25.
Define Avogadro Number.
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 x 10²³

Question 26.
Calculate the number of moles present in 60 g of ethane.
Answer:
No. of moles = = \(\frac {W}{M}\)
Molar Mass of ethane (C₂H₆) = 24 + 6 = 30
Number of moles in 60 g of ethane = \(\frac {60}{30}\) = 2 moles.

Question 27.
Calculate the equivalent mass of Copper. (Atomic mass of copper = 63.5)
Answer:
Equivalent mass = \(\frac {Atomic mass}{Valency}\)
Equivalent mass of Copper = \(\frac {63.5}{2}\) = 31.75 g eq^-1.

Question 28.
Calculate the equivalent mass of (i) Sulphate ion (ii) Phosphate ion.
Answer:
(i) Sulphate ion (SO₄^2-).
Equivalent mass of Sulphate =  = \(\frac {32 + 64}{2}\) = \(\frac {96}{2}\) = 48 g eq^-1

(ii) Phosphate ion (P0₄^3-)
Molar mass of Phosphate ion =  = \(\frac {31 + 64}{3}\) = \(\frac {95}{2}\) = 31.6 = 31.6g eq^-1

Question 29.
Calculate the equivalent mass of sulphuric acid.
Answer:
Sulphuric acid = H₂SO₄
Molar mass of Sulphuric acid = 2 + 32 + 64 = 96
Basicity of Sulphuric acid = 2
Equivalent mass of acid =  = \(\frac {96}{2}\) = 49g eq^-1

Question 30.
How many moles of hydrogen is required to produce 20 moles of ammonia?
Answer:
3H₂ + N₂ → 2NH₃
A per stoichiometric equation,
No. of moles of hydrogen required for 2 moles of ammonia 3 moles
No. of moles of hydrogen required for 20 moles of ammonia = \(\frac {3}{2}\) x 20 = 30 moles.

Question 31.
Calculate the amount of water produced by the combustion of 32 g of methane.
Answer:

As per stoichiometric equation,
16 g of methane produces 36 g of H₂O
∴ 32 g of methane will produce = \(\frac {36}{16}\) x 32 = 72 g of water.

Question 32.
How much volume of Carbon dioxide is produced when 25 g of calcium carbonate is heated completely under standard conditions?
Answer:

100 g of CaCO₃ produces 22.4 L of CO₂.
∴ 25 g of CaCO₃ will produce = \(\frac {22.4}{100}\) x 25 = 5.6 L of CO₂.

Question 33.
How much volume of chlorine is required to prepare 89.6 L of HCl gas at STP?
Answer:

2 x 22.4 L of HCl is produced by 22.4 L of Cl₂.
∴ 89.6 L of HCl will be produced by = img = 89.6 L = 44.8 L of chlorine.

Question 34.
What is meant by limiting reagent?
Answer:
A large excess of one reactant is supplied to ensure the more expensive reactant is completely converted to the desired product. The reactant used up first in a reaction is called the limiting reagent.

Question 35.
On the formation of SF₆ by the direct combination of S and F₂, which is the limiting reagent? Prove it.
Answer:
SF₆ is formed by burning Sulphur in an atmosphere of Fluorine. Suppose 3 moles of S is allowed to react with 12 moles of Fluorine.
S_(l) +3F_2(g) → SF_6(g)
As per the stoichiometric reaction, one mole of S reacts with 3 moles of fluorine to complete the reaction. Similarly, 3 moles of S requires only 9 moles of fluorine.
∴ It is understood that the limiting reagent is Sulphur and the excess reagent is Fluorine.

Question 36.
Mention any 4 redox reaction that takes place in our daily life.
Answer:

1. Burning of cooking gas, wood
2. Rusting of iron articles
3. Electroplating
4. Galvanic and electrolytic cells

Question 37.
Calculate the oxidation number of underlined elements in the following.

1. KMnO₄
2. Cr₂O₇^2-

Answer:
1. KMnO₄
1(+1) + x + 4 (-2) = 0
x – 7 = 0
∴ x = + 7
Oxidation state of Mn = +7.

2. Cr₂O₇^2-
2x + 7(-2) = -2
2x – 14 = – 2
2x = +l2
∴ x = + 6
Oxidation state of Cr = +6.

Question 38.
If 10 volumes of H₂ gas react with 5 volumes of O₂ gas, how many volumes of water vapour would be produced?
Answer:

Thus 2 volumes of H₂ reacts with 1 volume of O₂ to produce 2 volumes of H₂O_(g).
10 volumes of H₂ would react with 5 volumes of O₂ to produce 10 volumes of H₂O_(g)
Thus 10 volumes of H₂O will be produced.

Question 39.
Which one of the following will have largest number of atoms?

1. 1 g of Au(s)
2. 1 g of Na(s)
3. 1 g of Li(s)
4. 1 g of Cl₂ (g)

Answer:
1. Molar mass of Au = 197 g mol^-1.
No: of atoms in 1 g of Au = \(\frac {1}{197}\) x 6.023 x 10²³

2. Molar mass of Na = 23 g mol^-1.
No of atoms in 1 g of Na = \(\frac {1}{23}\) x 6.023 x 10²³

3. Molar mass of Li = 7 g mol^-1
No. of atoms in 1 g of Li = \(\frac {1}{7}\) x 6.023 x 10²³

4.  Molar mass of Cl₂ = 35.5 g mol^-1
No. of atoms in 1 g of Cl₂ = \(\frac {1}{35.5}\) x 6.023 x 10²³

Comparing the number of atoms, the largest number of atoms will be present in 1 g of Li. Since the mass is same in each case, the element with the lowest molar mass would have the largest number of atoms.
∴ Li with lowest molar mass would have the largest number of atoms.

Question 40.
What will be the mass of one ¹²C atom in g?
Answer:
Molar mass of ¹²C = 12.00 g mol^-1.
∴ Mass of 6.023 x 10²³ carbon atom = 12.0 g
∴ Mass of 1 carbon atom = \(\frac{12}{6.023 \times 10^{23}}\) = 1.992 x 10 g.

Question 41.
Justify the following reaction is a redox reaction.
Answer:

In the above reaction, oxygen is removed from CuO. So CuO gets reduced. Oxygen is added to H₂ to form water. So H₂ gets oxidised. i.e. In CuO, oxidation number of Cu +2 is reduced to O whereas in H₂, oxidation of H₂ O is increased to + 1. So the above reaction is a redox reaction.

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 3 – Mark Questions

Question 1.
Distinguish among the different physical states of matter.
Answer:
Differences among three physical states of matter (solid, liquid and gas) are as follows

S.No.	Properties	Solid	Liquid	Gas
1.	Volume	Definite	Definite	No definite
2.	Shape	Definite	No definite	No definite
3.	Molecular arrangement	Very closely packed	Loosely packed	Very loosely packed
4.	Freedom of movement	Not much freedom	Move around and better than solid	Move easy and fast
5.	Compressibility	Non compressible	Less compressible	Easily compressible

Question 2.
Define equivalent mass of a salt.
Answer:
Equivalent mass of a salt:
It is defined as the number of parts by mass of the salt that is produced by the neutralization of one equivalent of an acid by a base. Therefore the equivalent mass of the salt is equal to its molar mass.

Question 3.
How much copper can be obtained from 100 g of anhydrous copper sulphate?
Answer:
Anhydrous copper sulphate = CuSO₄
Molecular mass of CuSO₄ = 63.5 + 32 + (16 x 4)
= 63.5 + 32 + 64
= 159.5 g
159.5 g of CuSO₄ contains 63.5 g of copper.
∴ 100 g of CuSO₄ contains \(\frac{6.35}{159.5}\) x 100 = 0.39811 x 100 = 39.81 g of Copper.

Question 4.
Calculate the equivalent mass of hydrated ferrous sulphate.
Answer:
Hydrated ferrous sulphate = FeSO₄.7H₂O
Ferrous sulphate – Reducing agent
Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation.

16 parts by mass of oxygen oxidised 304 g of FeSO₄.
8 parts by mass of oxygen will oxidise \(\frac{304}{16}\) x 8 parts by mass of FeSO₄.
= 152
Equivalent mass of Ferrous sulphate (Anhydrous) = 152
Equivalent mass of crystalline Ferrous sulphate FeSO₄ .7H₂O = 152 + 126 = 278

Question 5.
Give difference between empirical and molecular formula
Answer:
Empirical Formula:

• Empirical formula is the simplest formula.
• It shows the ratio of number of atoms of different elements in one molecule of the compound.
• It is calculated from the percentage of composition of the various elements in one molecule.
• For example, Empirical formula of Benzene = CH.

Molecular Formula:

• Molecular Formula is the actual formula.
• It shows the actual number of different types of atoms present in one molecule of the compound.
• It is calculated from the Empirical formula Molecular Formula = (Empirical formula)_n
• Molecular formula of Benzene = C₆H₆.

Question 6.
A sample of hydrated copper sulphate is heated to drive off the water of crystallization, cooled and reweighed 0.869 g of CuSO4 aH₂O gave a residue of 0.556 g. Find the molecular formula of hydrated copper sulphate.
Answer:
0.869 g of CuSO₄.aH₂O gave a residue of 0.556 g of Anhydrous CuSO₄.
∴ Weight of a H₂O molecule = 0.869 – 0.556 = 0.313 g
Molecular weight of H₂O = (1 x 2) + 16 = 2 + 16 = 18
No. of moles of water = \(\frac{Mass}{Molecular mass}\)
CuSO₄.5H₂O – Molecular mass = 63.5 + 32 + 64 + 90 = 249.5 g
249.5 g of CuSO₄.5H₂O on heating gives 159.5 g of CuSO₄.
0.869 g of CuSO₄.aH₂O on heating gives = \(\frac{159.5}{249.5}\) x 0.869
= 0.556 g of anhydrous CuSO₄ ∴ a = 5
The molecular formula of hydrated copper sulphate = CuSO₄.5H₂O

Question 7.
Balance by oxidation number method: Mg + HNO₃ → Mg(NO₃)₂ + NO₂ + H₂O
Step – 1

Step – 2
Mg + 2HNO₃ -» Mg(NO₃)₂ + NO₂ + H₂O
Step – 3
To balance the number of oxygen atoms and hydrogen atoms 2HNO₃ is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + H₂O
Step – 4.
To balance the number of hydrogen atoms, the H₂O molecule is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + 2H₂O

Question 8.
Explain about the classification of matter.
Answer:

Question 9.
Calculate the mass of the following atoms in amu,
(a) Helium (mass of He = 6.641 x 10^-24 g)
(b) Silver (mass of Ag = 1.790 x 10^-22 g)
1 amu = 1.66056 x 10^-24
Answer:
(a) The mass of Helium atom in amu = \(\frac{6.641 \times 10^{-24}}{1.66056 \times 10^{-24}}\) = 3.9992 amu.
(b) The mass of Silver atom in amu = \(\frac{1.790 \times 10^{-22}}{1.66056 \times 10^{-24}}\) = 107.79 amu.

Question 10.
Calculate the number of atoms present in 1 Kg of gold.
Answer:
The atomic mass of Gold = 197 g mol^-1.
197 g of gold contains 6.023 x 10²³ atoms of gold.
∴ 1000 g of gold will contain = \(\frac{1000 \times 6.023 \times 10^{23}}{197}\)
= 3.055 x 10²⁴ atoms of Gold.

Question 11.
Calculate the molar volume of 146 g of HCl gas and the number of molecules present in it.
Answer:
Molar mass of HCl = 36.5 g
The molar volume of 36.5 g (1 mole) of HCl = 2.24 x 10² m³.
∴ The volume of 146 g (4 moles) of HCl = \(\frac{2.24 \times 10^{-2}}{36.5}\) x 146
= 8.96 x 10 m³
No. of molecules in 146 g of HCl = 4 N
= 4 x Avogadro Number
= 4 x 6.023 x 10²³
= 24.092 x 10²³
= 2.4092 x 10²⁴ molecules.

Question 12.
Calculate the molar mass of 20 L of gas weighing 23.2 g at STP.
Molar mass =

Molar volume at STP = 2.24 x 10^-2 m³ = 22.4 L (or) 22400 cc.
Molar mass of the gas at STP = \(\frac{23.2 x 22.4}{20}\) = 25.984 g.

Question 13.
0.6 g of a metal gives on oxidation 1 g of its oxide. Calculate its equivalent mass.
Answer:
Mass of metal = 0.6
Mass of metal oxide = 1 g
Mass of oxygen = 1 – 0.6 = 0.4 g
0.4 g of oxygen combines with 0.6 g of metal.
∴ 8 g of oxygen will combine with = \(\frac{0.6}{0.4}\) x 8
Equivalent mass of the metal = 12 g eq^-1.

Question 14.
How would you calculate the equivalent mass of anhydrous oxalic acid and hydrated oxalic acid.
Answer:
In acid medium,

16 g of oxygen is used for oxidation of 90 g of oxalic acid.
∴ 8 g of oxygen will oxidize = \(\frac{90}{16}\) x 8 = 45 g eq^-1.
Equivalent mass of Anhydrous oxalic acid = 45 g eq^-1

= \(\frac{126}{2}\) = 63 g eq^-1.

Question 15.
A compound on decomposition in the laboratory produces 24.5 g of nitrogen and 70 g of oxygen. Calculate the empirical formula of the compound.
Answer:

the empirical formula is N₂O₅

Question 16.
What is the steps involve in the calculation of molecular formula from empirical formula?
Answer:
Molecular mass and empirical formula are used to deduce molecular formula of the compound.
Steps to calculate molecular formula:

• if Empirical formula is found out from the percentage composition of elements
• Empirical formula mass can be found from the empirical formula
• Molecular mass is found out from the given data
• Molecular formula = (Empirical formula)_n
• where, n =

Question 17.
What is combination reaction? Give example.
When two or more substances combine to form a single substance, the reactions are called combination reactions.
A + B → C
Example:
2 Mg + O₂ → 2MgO

Question 18.
What is decomposition reaction? Give two examples.
Answer:
Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
AB → A + B
Example:
2KCO₃ → 2KCl + O₂
PCl₅ → PCl₃ + Cl₂

Question 19.
What is displacement reactions? Give its types. Explain with example.
Answer:
The reactions in which one ion or atom in a compound is replaced (or substituted) by an ion or atom of the other element are called displacement reactions.
AB + C → AC + B

Example:
Metal displacement
CuSO₄ + Zn → ZnSO₄ + Cu

Example:
Non-metal displacement
2KBr + Cl₂ → 2KCl + Br₂

Question 20.
What is disproportionation reactions? Give example.
Answer:
The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
Example:
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
2HCHO + H₂O → CH₃OH + HCOOH

Question 21.
What are competitive electron transfer reaction? Give example.
Answer:
These are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
Example:
Zn releases electrons to Cu and Cu releases electrons to Silver and so on.
Zn_(s) + Cu²⁺ → Zn²⁺_(aq) + Cu_(s) (Here Zn – oxidised; Cu²⁺ – reduced)
Cu_(s) + 2Ag⁺ → Cu²⁺_(aq) + 2Ag_(s) (Here Cu – oxidised; Ag⁺ – reduced)

Question 22.
Balance the following equation using oxidation number method.
Answer:
1. S + HNO₃ → H₂SO₄ + NO₂ + H₂O

2. S + 6HNO₃ → H₂SO₄ + NO₂ + H₂O
3. Balance the equation (except O and H)
S + 6HNO₃ → H₂SO₄ + 6NO₂ + H₂O
4. Balance O atoms by adding 2H₂O
5 + 6HNO₃ → H₂SO₄ + 6NO₂ + 2H₂O

Question 23.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.
Answer:

the empirical formula is Fe₂O₃

Question 24.
In three moles of ethane (C₂H₆) calculate the following:

1. Number of moles of carbon atoms.
2. Number of moles of hydrogen atoms.
3. Number of molecules of ethane.

Answer:

(1) 1 mole of C₂H₆ contains 2 moles of Carbon atoms.
∴ 3 moles of C₂H₆ will have 6 moles of Carbon atoms.
(2) 1 mole of C₂H₆ contains 6 moles of Hydrogen atoms.
∴ 3 moles of C₂H₆ will have 18 moles of Hydrogen atoms.
(3) 1 mole of C₂H₆ contains 6.023 x 10²³ number of molecules.
∴ 3 moles of C₂H₆ will contain 3 x 6.023 x 10²³molecules.

Question 25.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction.
4HCl_(aq) + MnO_2(s) → 2H₂O_(l) + MnCl_2(aq) + Cl_2(aq)
How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 g).
Answer:
1 mole of MnO₂ = 55 + 32 = 87 g.
87 g of MnO₂ reacts with 4 moles of HCl. i.e. = 4 x 36.5 = 146 g of HCl.
∴ 5 g of MnO₂ will react with \(\frac{146}{87}\) x 5.0 = 8.40 g.

Question 26.
The density of water at room temperature is 1.0 g/ml. How many molecules are there in a drop of water if its volume is 0.05 ml?
Answer:
Volume of drop of water = 0.05 ml
Mass of a drop of water = Volume x Density
= 0.05 ml x 1.0 g / ml.
= 0.05 g
Molar mass of water (H₂O) = 18 g
18 g of water = 1 mole
0. 05 g of water = \(\frac{1}{18}\) x 0.05 = 0.0028 mol.
No. of molecules present in one mole of water = 6.023 x 10²³
No. of molecules present in 0.0028 mole of water = \(\frac{6.023 \times 10^{23} \times 0.0028}{1}\) = 1.68 x 10²¹ water molecules

Question 27.
Balance the following equation by oxidation number method. MnO₄^– + Fe²⁺ → Mn ²⁺ + Fe³⁺ (Acidic medium)
Answer:

MnO₄^– + 5e^– → Mn²⁺ ……….(1)
Fe²⁺ → Fe³⁺ + e^– ……….(2)

to balance O and H atoms H₂O and H⁺ are added.
MnO₄^– + 5Fe²⁺ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 5-Mark Questions

I. Answer the following questions in detail
Question 1.
Define the following (a) equivalent mass of an acid (b) equivalent mass of a base (c) equivalent mass of an oxidising agent (cl) equivalent mass of a reducing agent.
Answer:
(a) Equivalent mass of an acid:
Equivalent mass of an acid is the number of parts by mass of the acid which contains 1.008 part by mass of replaceable hydrogen atom.
Equivalent mass of an acid =

(b) Equivalent mass of a base:
It is defined as the number of parts by mass of the base which contains one replaceable hydroxyl ion or which completely neutralizes one gram equivalent of an acid.
Equivalent mass of a base =

(c) Equivalent mass of an oxidising agent:
It is defined as the number of parts by mass of an oxidising agent which can furnish 8 parts by mass of oxygen for oxidation.

(d) Equivalent mass of a reducing agent:
It is defined as the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or with one equivalent of any oxidising agent.

Question 2.
Calculate the percentage composition of the elements present in lead nitrate. How many Kg of 02 can be obtained from 50 kg of 70% pure lead nitrate?
Answer:
Lead nitrate = Pb (NO₃)₂
Molecular mass of lead nitrate = 207 + (14 x 2) + (16 x 6)
= 207 + 28 + 96 = 331 g / mol.
331 g of lead nitrate contains 96 g of oxygen.
∴ 50 x 10³ g of lead nitrate will contain \(\frac {96}{331}\) x 50 x 10³
= 14501.5 g
= 14.501 Kg of oxygen.
100 % pure lead nitrate contains 14.501 Kg of oxygen.
70 % pure lead nitrate will contain = \(\frac {14.501}{100}\) x 70 = 10.15 Kg of oxygen.
.’. 70 % pure lead nitrate will contain 10.15 Kg of oxygen.

Question 3.
Determine the empirical formula of a compound containing K = 24.15%, Mn = 34.77% and rest is oxygen.
Answer:

empirical formula of a compound = KMnO₄

Question 4.
Write the steps to be followed for writing empirical formula.
Answer:
Empirical formula shows the ratio of number of atoms of different elements in one molecule of the compound.
Steps for finding the Empirical formula:
The percentage of the elements in the compound is determined by suitable methods and from the data collected; the empirical formula is determined by the following steps.

1. Divide the percentage of each element by its atomic mass. This will give the relative number of atoms of various elements present in the compound.
2. Divide the atom value obtained in the above step by the smallest of them so as to get a simple ratio of atoms of various elements.
3. Multiply the figures so obtained, by a suitable integer if necessary in order to obtain whole number ratio.
4. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.
5. Percentage of Oxygen = 100 – Sum of the percentage masses of all the given elements.

Question 5.
An organic compound was found to contain carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour density was found to be 29.5. What is the molecular formula of the compound?
Answer:

Empirical formula = C₂H₅NO
Molecular mass = 2 x vapour density = 2 x 29.5 = 59.0
Empirical formula mass of C₄H₅NO = 24 + 5 + 14 + 16 = 59
n =  = \(\frac {59}{59}\) = 1
Empirical formula mass 59
.’. Molecular formula = (Empirical formula)_n
= (C₂H₅NO)₁
Molecular formula = C₂H₅NO

Question 6.
Calculate the empirical and molecular formula of a compound containing 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75.
Answer:

Empirical formula = C₂H₃O₃
Empirical formula mass = 24 + 3 + 48 = 75
Molecular mass = 2 x Vapour density = 2 x 75 = 150
n =  = \(\frac {150}{75}\) = 2
Molecular formula = (Empirical formula)_n
Molecular formula = C₂H₃O₃ x 2
Molecular formula = C₄H₆O₆

Question 7.
Explain the different types of redox reactions with example.
Answer:
Redox reactions are classified into the following types:
(1) Combination reactions:
When two or more substances combine to form a single substance, the reactions are called combination reactions.
Example:
2Mg + O₂ → 2MgO

(2) Decomposition reactions:
Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
Example:
2KClO₃ → 2KCl + 3O₂

(3) Displacement reactions:
The reactions in which one ion or atom in a compound is replaced by an ion or atom of the other element are called displacement reactions.
Example:
CuSO₄ + Zn → ZnSO₄ + Cu

(4) Disproportionation reactions:
The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
Example:
2HCHO + H₂O → CH₃OH + HCOOH

(5) Competitive Electron transfer reactions:
These are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
Example:
Zn_(s) + Cu²⁺ → Zn²⁺_(aq) + Cu_(s)  Here Zn – oxidised; Cu²⁺ reduced

Question 8.
Write the steps to be followed while balancing redox equation by oxidation number method.
Answer:
Oxidation number method:
This method is based on the fact that
Number of electrons lost by atoms = Number of electrons gained by atoms

Steps to be followed while balancing Redox reactions by Oxidation Number method:

1. Write skeleton equation representing redox reaction
2. Write the oxidation number of atoms undergoing oxidation and reduction.
3. Calculate the increase or decrease in oxidation numbers per atom.
4. Make increase in oxidation number equal to decrease in oxidation number by multiplying the formula of oxidant and reductant by suitable numbers.
5. Balance the equation atomically on both sides except O and H atoms.
6. Balance oxygen atoms by adding required number of water molecules to the side deficient
   in oxygen atoms.
7. Add required number of H+ ions to the side deficient in hydrogen atom if the reaction is in acidic medium.
8. For reactions in basic medium, add H₂O molecules to the side deficient in hydrogen atoms and simultaneously add equal number of OH ions on the other side of the equation.
9. Finally balance the equation by cancelling common species present on both sides of the equation.

Question 9.
Balance the following equation by oxidation number method:
Answer:
K₂Cr₂O₇ + KCl + H₂SO₄ → KHSO₄ + CrO₂Cl ₂ + H₂O
Step – 1

Step – 2
K₂Cr₂O₇ + KCl + H₂SO₄ → KHSO₄ + 2CrO₂C₂ + H₂O

Step 3.
To balance Cl atom, KC1 is multiplied by 4
K₂Cr₂O₇ + 4KCl + H₂SO₄ → 2CrO₂Cl₂ + KHSO₂ + H₂O

Step 4.
To balance K atom, KHSO₄ is multiplied by 6.
K2Cr2O₇ + 4KCl + H2SO₄ → 2CrO₂Cl₂ + 6KHSO₄+ H₂O

Step 5.
To balance O and H atoms, HSO₄ is multiplied by 6, H20 is multiplied by 3.
Answer:
K₂Cr₂O₇ + 4KCl + 6H₂SO₄ → 2CrO₂Cl₂ + 6KHSO₄ + 3H₂O

(2) P + HNO₃ → H₃PO₄ + NO₂ + H₂O
Step – 1

Step 2.
P + 5HNO₃ → H₃PO₄ + 5NO₂ + H₂O

(3) CuO + NH₃ → Cu + N₂ + H₂O
Step 1.

Step 2.
3CuO + 2NH₃ → 3Cu + N₂ + H₂O

Step 3.
To balance O and N, water is multiplied by 3.
3CuO + 2NH₃ → 3Cu + N₂ + 3H₂O

(4) Zn + HNO₃ →Zn(N0₃)₂ + NH₄NO₃ + H₂O
Step – 1.

Step 2.
4Zn + HNO₃ → 4Zn(NO₃)₃ + NH₄NO₂ + H₂O

Step 3.
To balance N, HNO₃ is multiplied by 10
4Zn + 10HNO3 → 4Zn(NO₃)₂ + NH₄NO₃ + H₃O

Step 4.
To balance oxygen, H₂O is multiplied by 3
4Zn + 10HNO₃ → 4Zn(NO₃)₂ + NH₄NO₃ + 3H₃O

Question 10.
Balance the following equation by ion-electron method In acidic medium.
Answer:
(i) S₂O₃^2- + I₂ → S₂O₄^2- + SO₂ + I^–
Oxidation half reaction:

Reduction half reaction:

To balance, SO₂ is added on RHS of the equation.
S₂O₃^2- + I₂ → S₂O₄^2- + 2I^– + SO₂
To balance oxygen atom, S₂O₃^2- and SO₂ is multiplied by 2.
2S₂O₃^2- + I₂ → S₂O₄^2- + 2I – + 2SO₂

(2) Sb³⁺ + MnO₄ → Sb⁵⁺ + Mn²⁺
Oxidation half reaction:
Sb³⁺ → Sb⁵⁺ + 2e ^– ……….(1)
Reduction half reaction:

In equation (2), H₂O is added on L.HS to balance oxygen atom.
MnO₂ + 5e^– → Mn²⁺ + 4H₂O ………(3)
To balance Hydrogen atoms, H’ is added on RHS.
MnO₄^– + 5e^– + 8H⁺ → Mn²⁺ + 4H₂O ………(4)
Equation (4) is multiplied by 2 and equation (I) is multiplied by 5 to equalise the electrons gained and electrons lost.

(3) MnO₄^– + I^– → MnO₂ + I₂
Oxidation half reaction:
2I^– + 2e^– → I^– ………..(1)
Reduction half reaction:
MnO₄^–  → MnO₂ + e^– ………..(2)                                    
Equation (1)is multiplied by 3

Equation (2)is multiplied by 2.

To balance oxygen and hydrogen atoms, H⁺ is added on RHS and H₂O is added on LHS.
2MnO₄ + 6I^– + 4H⁺ → 2MnO₂ + 2I₂ + 2H₂O

In acidic medium
(4) MnO₄^– + Fe²⁺ → Mn²⁺ + Fe³⁺
Oxidation half reaction:
Fe²⁺ → Fe³⁺ + e^– ……..(1)
Reduction half reaction:

To balance oxygen, H⁺ is added on RHS and H₂O is added on LHS.
MnO₄^– + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

(5) Cr(OH)₄^– + H₂O₂ → CrO₄^–
Oxidation half reaction:

Reduction half reaction:

To balance oxygen and hydrogen atoms, OH and H₂O are added.
2Cr(OH)₄ + 3H₂O₂ + 20H^– → 2CrO₂^– + 8H₂O

Question 11.
(a) Define equivalent mass of an oxidising agent.
(b) How would you calculate the equivalent mass of potassium permanganate?
(a) The equivalent mass of an oxidizing agent is the number of parts by mass which can furnish 8 parts by mass of oxygen for oxidation.
(b) Potassium permanganate is an oxidizing agent.

80 parts by mass of oxygen are given by 316 g of KMnO₄.
.’. 8 parts by mass of oxygen will be furnished by \(\frac {316}{80}\) x 8 = 31.6
Equivalent mass of KMnO₄ = 31.6 g eq^-1.

Question 12.
(a) Define equivalent mass of an reducing agent.
(b) How would you determine the equivalent mass of Ferrous sulphate?
Answer:
(a) The equivalent mass of a reducing agent is the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or one equivalent of any oxidising agent.
(b) Ferrous sulphate is a reducing agent.

16 parts by mass of oxygen oxidised 304 parts by mass of FeSO₄.
∴ 8 parts by mass of oxygen will oxidise \(\frac {316}{80}\) x 8 parts by mass of ferrous sulphate = 152 .
The equivalent mass of ferrous sulphate (anhydrous) = 152.
The equivalent mass of crystalline ferrous sulphate is (FeSO₄.7H₂O) = 152 + 126 = 278
The equivalent mass of crystalline ferrous sulphate = 278.

Question 13.
A compound on analysis gave the following percentage composition: C = 24.47%, H = 4.07 %, Cl = 7 1.65%. Find out its empirical formula.
Answer:

the empirical formula is CH₂Cl.

Question 14.
A laboratory analysis of an organic compound gives the following mass percentage composition: C = 60%, H = 4.48% and remaining oxygen.
Answer:

∴ the empirical formula is C₉H₈O₄.

Question 15.
An insecticide has the following percentage composition by mass: 47.5% C, 2.54% H, and 50.0% Cl. Determine its empirical formula and molecular formulae. Molar mass of the substance is 354.5 g mol^-1
Answer:

The empirical formula is C₁₄H₉C₁₅.
Calculation of Molecular formula:
The empirical formula mass (C₁₄H₉C₁₅) = (14 X 12) + (9 X 1) + (5 X 35.5)
n =  = \(\frac {354.5}{354.5}\) = 1
Empirical formula mass 354.5
Molecular formula = (Empirical formu1a)_n
= (C₁₄H₉Cl₅)₁
∴ Molecular formula = C₁₄H₉Cl₅

Question 16.
An organic fruit smelling compound on analysis has the following composition by mass: C = 54.54%, H = 9.09%, O = 36.36%. Find out the molecular formula of the compound. The vapour density of the compound was found to be 44.
Answer:

∴ The empirical formula is C₂H₄O
Theempirical formula mass(C₂H₄O) = (12 x 2)+(1 x 4)+(16 x 1) = 44
Molecular mass = 2 x Vapour density = 2 x 44 = 88
n =  = \(\frac {88}{44}\) = 2
Molecular formula = (Empirical formula)_n
= (C₂H₄O)₂
∴ Molecular formula = C₂H₄O₂

Question 17.
Calculate the percentage composition of the elements present in magnesium carbonate. How many Kg of CO₂ can be obtained from 100 Kg of is 90% pure magnesium carbonate.
Molar mass of MgCO₃ = 84.32 g mol^-1
Percentage of Mg = \(\frac {24}{84.32}\) x 100 = 28.46%
Percentage of C = \(\frac {12}{84.32}\) x 100 = 14.23%
Percentage of O₃ = \(\frac {48}{84.32}\) x 100 = 57.0%

84.32 g of 100% pure MgCO₃ gives 44g of CO₂
∴ 100 x 10³g of 100% pure MgCO₃ gives = \(\frac {44}{84.32}\) x 100 x 10³
= 52.182 x 10³ g CO₂
100% pure MgCO₃ gives 52.182 x 10³ g CO₂
∴ 90% pure MgCO₃ will give \(\frac{52.182 \times 10^{3}}{100}\) x 90 = 46963.8 g CO₂

Question 18.
Urea is prepared by the reaction between ammonia and carbon dioxide.

In one process, 637.2 g of NH₃ are allowed to react with 1142 g of CO₂.
(a) Which of the two reactants is the limiting reagent?
(b) Calculate the mass of(NH₄)₂CO formed.
(c) how much of the excess reagent in grams is left at the end of the reaction?

No. of moles of ammonia = \(\frac{637.2}{17}\) = 37.45 mole
No. of moles of CO₂ = \(\frac{1142}{44}\) = 25.95 moles
As per the balanced equation, one mole of CO₂ requires 2 moles of ammonia.
∴ No. of moles of NH₃ required to react with 25.95 moles of CO₂ is = \(\frac {2}{1}\) x 25.95 = 51.90 moles.
∴ 37.45 moles of NH₃ is not enough to completely react with CO₂ (25.95 moles).
Hence, NH₃ must be the limiting reagent, and CO₂ is excess reagent.

(b) 2 moles of ammonia produce 1 mole of urea.
∴ Limiting reagent 37.45 moles of NH₃ can produce \(\frac {1}{2}\) x 37.45 moles of urea.
= 18.725 moles of urea.
∴ The mass of 18.725 moles of urea = No. of moles x Molar mass
= 18.725 x 60
= 1123.5 g of urea.

(c) 2 moles of ammonia requires 1 mole of CO₂.
∴ Limiting reagent 37.45 moles of NH₃ will require x 37.45 moles of CO₂.
= 18.725 moles of CO₂.
∴ No. of moles of the excess reagent (CO₂) left = 25.95 – 18.725 = 7.225
The mass of the excess reagent (CO₂) left = 7.225 x 44 = 317.9 g CO₂.

Question 19.
(a) Define oxidation number.
(b) What are the rules used to assign oxidation number?
Answer:
(a) Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely.
(b) Rules to assign oxidation number:

• Oxidation number of a substance in its elementary state is equal to zero (H₂, Br₂, Na)
• Oxidation number of a mono-atomtic ion is equal to the charge on the Ton (Na₄ =+ 1,
• Oxidation number of hydrogen in a compound is +1 (except hydrides).
• Oxidation number of hydrogen in metal hydrides is -1 (NaH, CaH₂).
• Oxidation number of oxygen in a compound is -2 (except OF₂ and peroxides).
• Oxidation number of oxygen in peroxides is -1 (He,, Na»,) = I.
• Oxidation number of oxygen in fluorinated compounds is either +1 or +2.(OF₂ = +2, O,F₂ + 1).
• Fluorine has an oxidation number -1 in all ìts compounds.
• The sum of the oxidation number of all the atoms in neutral molecules is equal to Zero.
• For all ions, the sum of the oxidation number of all atoms is equal to the charge of the ion.

Question 20.
Balance the following equation by oxidation number method.
C₆H₆ + O₂ → CO₂ + H₂O

(ii) Balance the changes in O.N. by multiplying the oxidant and reductant by suitable numbers
2 C₆H₆ + 15 O₂ → CO₂ + H₂O

(iii) Balance the equation atomically (except O and H).
2C₆H₆ + 15 O₂ → 12 CO₂ + H₂O

(iv) Balance O atoms by adding one HzO molecule to the RHS for making the number of molecules of H₂O to be 6.
2C₆H₆ + 15 O₂ → 12 CO₂ + 6H₂O

Question 21.
Balance the following equation by oxidation number method.
KMnO₄ + HCl → KCl + MnCl₂ + H₂O + Cl₂

(ii) 2KMnO₄ + 1OHCl → KCl + MnCl₂ + H₂O + Cl₂

(iii) Balance the equation atomically (except O and H).
2KMnO₄ + 1OHCl → 2KCl + 2MnCl₂, + H₂O + Cl₂

(iv) Balance chlorine atoms by adding HC1 and multiplying Cl₂ by 5.
2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + H₂O + 5Cl₂

(v) To balance O and H, H₂O is multiplied by 8.
2KMnO₄  + I6HCl → 2KCl + 2MnCl₄ + 8H₂O + 5Cl₂

Question 22.
Balance the following equation by oxidation number method.
KMnO₄ + FeSO₄ + H2S0₄ → K2SO₄ + MnSO₄ + Fe2(S0₄)₃ + H₂O
Answer:

(ii) 2KMnO₄ + 10FeSO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + Fe2(SO₄)₃+ H₂O

(iii) Balance the equation atomically (except O and H) and sulphate ions.
2KMnO₄ + 1OFeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₄(SO₄)₃ + H₂O

(iv) Balance O atoms by multiplying H₂O by 8.
2KMnO₄ + 1OFeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O

Question 23.
Balancing of molecular equation in alkaline medium.
MnO₂ + O₂ + KOH → K₂MnO₄ + H₂O

(ii) Balance the changes in O, N, by multiplying the oxidant and reductant by suitable numbers.
4MnO₂ + 2O₂+ KOH → K₂MnO₄ + H₂O

(iii) Balance the equation atomically (except O and H).
4MnO₂ + 2O₂ + KOH → K₂ MnO₄ + H₂O

(iv) Balance oxygen and hydrogen atoms by multiplying H20 by 4.
4MnO₂ + 2O₂ + 8KOH → 4K₂MnO₄ + 4H₂O

Question 24.
Explain the steps involved in ion-electron method for balancing redox reaction.
Answer:
Ion electron method makes use of the Half reactions. Steps involved in this method are,

1. Write the equation in the net ionic form without attempting to balance it.
2. Write and locate the oxidation number of atoms undergoing oxidation and reduction from the knowledge of calculation of oxidation number.
3. Write two half reactions showing oxidation and reduction separately.
4. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.
5. Add required number of H⁺ ions to the side deficient in hydrogen atom if the reaction is in acidic medium
6. Add electrons to whichever side is necessary to make up the difference in oxidation number.
7. Add the two half reactions. The resulting equation is a net balanced equation.
8. For reactions in basic medium, add H₂O and hydrogen ion to balance H and O.
9. Finally balance the equation by cancelling common species present on both sides of the equation.

Question 25.
Write balanced equation for the oxidation of Ferrous ions to Ferric ions by permanganate ions in acid solution. The permanganate ion forms Mn2+ ions under these conditions.
Answer:
Net ionic reaction:
MnO₄^– + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺
Oxidation half reaction:
Fe²⁺ → Fe³⁺ + e^– ……….(1)
Reduction half reaction:
MnO₄^– + 5e^– → Mn²⁺ ……….(2)

To balance O and H, H⁺ and H₂O are added.
MnO₄^– + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ +4H₂O

Question 26.
A flask A contains 0.5 mole of oxygen gas. Another flask B contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms.
Answer:
Flask A:
1 mole of oxygen gas = 6.023 x 10²³ molecules
∴ 0.5 mole of oxygen gas = 6.023 x 10²³ x 0.5 molecules
The number of atoms in flask A = 6.023 x 10²³ x 0.5 x 2 = 6.023 x 10²³ atoms.

Flask B:
1 mole of ozone gas = 6.023 x 10²³ molecules
0. 4 mole of ozone gas = 6.023 x 10²³ x 0.4 molecules
The number of atoms in flask B = 6.023 x 10²³ x 0.4 x 3 = 7.227 x 10²³ atoms.
∴ Flask B contains a great number of oxygen atoms as compared to flask A.

Question 27.
(a) Formulate possible compounds of ‘Cl’ in its oxidation state is:
0, – 1, + l, + 3,+ 5, + 7

(b) H₂O₂ act as an oxidising agent as well as reducing agent where as O₃ act as only oxidizing agent. Prove it.
(a) (1) Cl oxidation number O in Cl₂.
(2) Cl oxidation number -1 in HCl.
(3) Cl oxidation number +3 in HCl O₂.
(4) Cl oxidation number +5 in KClO₃.
(5) Cl oxidation number +7 in C1₂O₇.

(b) In H₂O₂, oxidation number of oxygen is -1 and it can vary from O to -2 (+ 2 is possible in OF₂). The oxidation number can decrease or increase, because of this H₂O₂ can act both oxidising and reducing agent. Ozone (O₃ ) only acts as oxidising agent since it decomposes to give nascent oxygen.

Question 28.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H ion. Write a balanced ionic equation for the reaction.
The skeletal equation is:
Mn³⁺_(aq) → Mn²⁺_(aq) + MnO_2(aq) + H⁺_(aq)
Oxidation half equation:

Balance O.N. by adding electrons,
Mn³⁺_(aq) →  MnO_2(s) + e^–
Balance charge by adding 4H ions,
Mn³⁺_(aq) →  MnO_2(s) + 4H⁺_(aq) + e^–
Balance O atoms by adding 2H₂O
Mn³⁺_(aq) + 22H₂O_(l) → MnO_2(s) + 4H⁺_(aq) + e^– ……….(1)
Reduction half equation:

Balance ON. by adding electrons:
Mn³⁺_(aq) + e^– → Mn²⁺_(aq) …………(2)
Adding Equation (1) and (2), the balanced equation for the disproportionation reaction is
2MH³⁺_(aq) + 2H₂O_(l) → MnO_2(s) + Mn²⁺_(aq) + H⁺_(aq)

Question 29.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating with sulphur dioxide. Present a balanced equation for the reaction for this redox change taking place in water.
The skeletal equation is:
Cl_2(aq) + SO_2(aq) + H₂O_(l) → Cl^–_(aq) + SO₄^2-_(aq)
Reduction half equation:
Cl_2(aq) → Cl^2-_(aq)
Balance Cl atoms,

Balance O.N. by adding electrons
Cl_2(aq) + 2e^– → 2Cl + 2e^–
Oxidation half equation:

Balance O,N. by adding electrons
SO_2(aq) → SO₄^2-_(aq) + 2e^–
Balance charge by adding 4H⁺ ions:
SO_2(aq) → SO₄^2-_(aq) + 4H⁺_(aq) + 2e^–
Balance O atoms by adding 2H₂O
SO_2(aq) + 2H₂O_(l) → SO₄^2-_(aq) + 4H⁺_(aq) + 2e^–
Adding equation (1) and (2), we have,
Cl_2(aq) + SO_2(aq) + 2H₂O_(l) → Cl^–_2(aq) + SO₄^2-_(aq) + 4H⁺_(aq)
This represents the balanced redox reaction.

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Tamilnadu Samacheer Kalvi 9th Social Science Geography Solutions Chapter 5 Biosphere

Biosphere Textual Exercise

I. Fill in the blanks.

1. An area where animals, plants and micro organisms live and interact with one another is known as ………
2. ……… are also called Heterotrophs.
3. ……. is a system of interlocking and independent food chains.
4. ………. is an extensive large ecosystem.
5. The vegetative type commonly found in desert biomes is called …..
6. ……. is an aquatic biome that is found where fresh water and salt water mix.
Answers:
1. Habitat
2. Consumers
3. Food web
4. A biome
5. Xerophytes
6. Estuary

II. Choose the correct answer.

Biosphere Provides A Stable Ecosystem Give Reason Question 1.
The coldest biome on Earth is ……….
(a) Tundra
(b) Taiga
(c) Desert
(d) Oceans
Answer:
(a) Tundra

Biosphere Provides A Stable Ecosystem Question 2.
This is the smallest unit of biosphere.
(a) Ecosystems
(b) Biome
(c) Environment
(d) None of the above
Answer:
(a) Ecosystems

Biosphere 9th Class Social Question 3.
Nutrients are recycled in the atmosphere with the help of certain micro organisms, referred to as
(a) Producers
(b) Decomposers
(c) Consumers
(d) None of the above
Answer:
(b) Decomposers

Biosphere Lesson Class 9 Question 4.
To which climatic conditions are Xerophytic plants specifically adapted to?
(a) Saline and sandy
(b) Limited moisture availability
(c) Cold temperature
(d) Humid
Answer:
(a) Saline and sandy

Biosphere Class 9 Question 5.
Why is the usage of rainforest biomes for large scale agriculture unsustainable?
(a) because it is too wet
(b) because the temperature is too warm
(c) because the soil is too thin
(d) because the soil is poor
Answer:
(b) because the temperature is too warm

Questions 6-8 are assertion type questions.
Directions:
(A) Both assertion (A) and reason (R) are true; R explains A
(B) Both assertion (A) and reason (R) are true; R does not explain A
(C) A is true; R is false
(D) Both A and R are false
Answer:
(C) A is true; R is false

Biosphere Solutions Question 6.
(A): Heterotrophs do not produce their own food.
(R): They depend on autotrophs for their nourishment.
Answer:
(C) A is true; R is false

Biosphere Provides A Stable Ecosystem. Question 7.
(A): Hotspots are the regions characterised by numerous endemic plants and animal species
living in a vulnerable environment.
(R): To manage and focus on conservation work more effectively, researchers identified hotspots.
Answer:
(A) Both assertion (A) and reason (R) are true; R explains A

9th Class Social Biosphere Lesson Question 8.
(A): The number of gorillas in Africa has plummeted by 60% in the past twenty years.
(R): Non intervention of human beings in the forest areas.
Answer:
(D) Both A and R are false

III. Answer the following in brief:

9th Class Social Biosphere Question 1.
What is Biosphere?
Answer:
Biosphere is a life supporting layer that exists on the earth’s surface. This layer on earth encompasses the Lithosphere, Hydrosphere and Atmosphere. It includes flora and fauna that thrive on or near the earth’s surface.

Samacheer Kalvi 9th Social Science Geography Chapter 1 Question 2.
What is an ecosystem?
Answer:
The place on earth where living organisms live and interact with one another and with their physical environment is called an ecosystem.

Question 3.
What does the term ‘biodiversity’ meant
Answer:
Biodiversity or biological diversity refers to a wide variety of living organisms (plants, animals and other micro organisms) which live in a habitat. It is highly influenced by topography, climate as well as human activities.

Question 4.
What is meant by loss of biodiversity?
Answer:
The extinction of species (flora and fauna) due to human and natural influences is called loss of biodiversity. The biodiversity loss has a great impact on mankind and also affects land, water, air etc. Habitat destruction due to deforestation, population explosion, pollution and global warming are the major cause for loss of biodiversity.

Question 5.
Mention the various terrestrial biomes.
Answer:
The major terrestrial biomes of the world are:
A. Tropical Forest Biomes
B. Tropical Savanna Biomes
C. Desert Biomes
D. Temperate Grassland Biomes
E. Tundra Biomes

IV. Distinguish between the following:

Question 1.
Producers and Decomposers.
Answer:

S.No.	Producers	Decomposers
(i)	They are self nourishing components of the ecosystem.	They are organisms that are incapable of preparing its own food.
(ii)	Producers are called Autotrophs. e.g. Plants, Algae, Bacteria, etc.	Decomposers are organisms that are incapable of preparing its own food. They are called Saprotrophs. e.g. Fungus, Mushrooms.

Question 2.
Terrestrial biomes and Aquatic biomes
Answer:

S.No.	Terrestrial biomes	Aquatic biomes
(i)	It is a group of living organisms that live and interact with one another on land.	It is a group of living organisms that live and interact with one another for nutrients and shelter.
(ii)	They are determined by temperature and rainfall.	These are influenced by a series of abiotic factors.
(iii)	Terrestrial biomes are Tropical forest biomes, Tropical Savanna biomes, Desert biomes, Temperate Grassland biomes and Tundra biomes.	Aquatic biomes are Fresh water biomes and Marine biomes.

Question 3.
Tropical vegetation and Desert vegetation
Answer:

S.No.	Tropical vegetation	Desert vegetation
(i)	Half of the World’s Tropical forests are found in Central and South America.	Deserts are found in the Western Margin of the continents.
(ii)	This biome extends between 10°N and 10°S of the equator.	This extends between 20°N and 30°S latitudes.
(iii)	This weather condition favours thick vegetative cover.	They possess special type of vegetation called xerophytes.
(iv)	People in this region sustain their livelihood through food gathering.	People in this region sustain their life through food gathering and hunting.
(v)	High annual rainfall and relatively constant temperature is likely	Annual rainfall is less than 25 c.m in these regions.

Question 4.
Savannas and Tundra
Answer:

S.No.	Savannas	Tundras
(i)	Tropical grasslands are generally found between tropical forests.	These vast lowlands are found where the ground remains frozen.
(ii)	Generally hot climate and dry. They also experience moderate to low rainfall.	Generally long severe winter and short cool summer.
(iii)	Found in the Sahel, South of Sahara in East Africa and in Australia.	Found in Greenland Arctic and Antarctic regions and Northern parts of Asia, Canada and Europe.
(iv)	Chief occupation is herding.	Hunting and fishing are their main occupations.

V. Give reasons for the following:

Question 1.
Producers are also called autotrophs.
Answer:
Producers are self nourishing components of the ecosystem. Hence they are called Autotrophs. They are found both on land and water, e.g. Plants, Algae, Bacteria etc.

Question 2.
Biosphere provides a stable ecosystem.
Answer:

1. Biosphere extends from the deep ocean trenches to lush rain forests.
2. People play an important role in maintaining the flow of energy in the biosphere.
3. There are places on earth that are both biologically rich and deeply threatened.
4. Hence Biosphere provides a stable ecosystem.

VI. Answer the following in a paragraph:

Question 1.
Explain the various components of ecosystem.
Answer:
An ecosystem is a community, where all living organisms live and interact with one another and also with their non-living environment such as land, soil, air, water etc. Ecosystems range in size from the smallest units (e.g. bark of a tree) that can sustain life to the global ecosystem or ecosphere. (e.g. Cropland, Pond ecosystem, Forest ecosystem, Desert ecosystem etc.). Biosphere harbours all ecosystems on the earth and sustains life forms including mankind.
Components of ecosystem:
An ecosystem consists of three basic components, namely
(A) Abiotic components
(B) Biotic components and
(C) Energy component

A) Abiotic Components: Abiotic components include the non-living, inorganic, physical and chemical factors in the environment, e.g. Land, Air,Water, Calcium, Iron etc.

B) Biotic Components: Biotic components include plants, animals and micro organisms. Biotic components can be classified into three categories:

• Producers are self nourishing components of the ecosystem. Hence they are called Autotrophs. They are found both on land and water, e.g. Plants, Algae, Bacteria etc.
• Consumers are those that depend on producers, directly or indirectly. Hence they are called Heterotrophs.

C) Energy Components: All organisms in the biosphere use energy to work and convert one form of energy into another. The Sun is the ultimate source of energy for the biosphere as a whole. The solar energy gets transformed into other forms of energy through the various components in the ecosystem. The producers, consumers and the decomposers contribute a lot to the energy flow in an ecosystem.

Question 2.
Write a paragraph on the functions of an ecosystem.
Answer:
Functions of an ecosystem: The living organisms form an interacting set of flora and fauna which are organized into trophic levels, food chains and food webs. The functioning of an ecosystem depends on the pattern of the energy flow, as it helps in the distribution and circulation of the organic and inorganic matter within an ecosystem. Energy flow generally takes place in a hierarchical order in an ecosystem through various levels. These levels are called trophic levels. The chain of transformation of energy from one group of organisms to another, through various trophic levels is called a food chain. A system of interlocking and interdependent food chains is called a food web.

Question 3.
Explain about the aquatic biomes on Earth.
Answer:
Aquatic biome is a group of living organisms that live and interact with one another and its aquatic environment for nutrients and shelter. Like terrestrial biomes, aquatic biomes are influenced by a series of abiotic factors. It is broadly classified as fresh water biomes and marine biomes.

A. Fresh water Biomes: It comprises lakes, ponds, rivers, streams, wetlands etc. It is influenced by various abiotic components such as the volume of water, water flow, composition of oxygen, temperature, etc. Humans rely on freshwater biomes for drinking water, crop irrigation, sanitation and industry. Water lily, lotus, duck weeds etc. are the common plants found here. Trout, salmon, turtles, crocodiles etc. are the animals found here.

B. Marine Biomes: They are the largest aquatic biomes on earth. They are continuous bodies of salt water and provide a wide range of habitats for marine plants and animals. Coral reefs are a second kind of marine biomes within the ocean. Estuaries, coastal areas where salt water and fresh water mix, form a third unique marine biome. As water provides maximum mobility to marine organisms, nutrients are circulated more quickly and efficiently here than the terrestrial biomes. Apart from animals, plants such as kelp, algae, phytoplankton etc. also grow in water. Aquatic biomes are not only important for plants and animals, but also for humans. Humans use aquatic biomes for water, food and leisure activities. Some of the threats and issues to aquatic biomes are overfishing, pollution and rise in sea level.

VII. Find out the dates for the following:

1. World Wild Life Day – …………
2. International Day of Forest – ………..
3. World Water Day – ………..
4. Earth Day – ………….
5. World Environment Day – June 5th
6. World Oceans Day – …………
Answers:
1. March 3rd
2. March 21st
3. March 22nd
4. April 22nd
6. June 8th

VIII. Map Study

Locate the following on the world outline map.

1. Priairies
2. Downs
3. Tundra Biomes
4. Equatorial Biomes

You can identify and locate the above mentioned grasslands with the help of an atlas.
Hint: Temperate grasslands are called differently in different parts of the world.
Prairies – North America
Downs – Australia and New Zealand
Tundra – Greenland, Arctic and Antarctic regions and Northern parts of Asia, Canada and Europe
Equatorial Biomes – Near the equator in Central and South America, parts of Africa and Asia

IX. Picture Study

Narrate the given food web of Arctic Tundra in your own words.

1. The flow of energy from the sun through an eco system can be illustrated in a food chain.
2. Within a food Chain there are produces. e.g., grasses, lichens, Arctic wild flower, etc.
3. Life in any biome includes producers (plants) and consumers (energy obtained from producers)
4. These consumers may be further classified as herbivores (plant eater) carnivores(meat eaters) (or) anniores (feed on combination of both animals and plants).

Biosphere Additional Questions

I. Fill in the blanks.

1. ……… are self nourishing components of the ecosystem.
2. All living things, large or small are grouped into ………
3. The ……. is the ultimate source of energy for the biosphere as a whole.
4. Decomposers live on dead and decaying plants and animals and are called ………
5. The extinction of species (flora and fauna) due to human and natural influences is called ………
6. The branch of science that deals with ecosystem is called ………
7. The temperate grasslands are called …… in North America.
8. ……….. is a fertile fresh water source found in deserts and semi-arid regions.
Answers:
1. Producers
2. species
3. sun
4. Saprotrophs
5. Lass of biodiversity
6. Ecology
7. Prairies
8. An Oasis

II. Choose the correct answer.

Question 1.
……….. the fourth sphere of the Earth.
(a) Lithosphere
(b) Hydrosphere
(c) Biosphere
(d) Atmosphere
Answer:
(c) Biosphere

Question 2.
………. are found both on land and water.
(a) Autotrophs
(b) Heterotrophs
(c) Saprotrophs
(d) None of the above
Answer:
(a) Autotrophs

Question 3.
Consumers are those that depend on directly (or) indirectly are called ………..
(a) Saprotrophs
(b) Heterotrophs
(c) Autotrophs
(d) None of the above
Answer:
(b) Heterotrophs

Question 4.
A group of living organisms that live and interact with One another on land is ………
(a) Tropical Savanna Biomes
(b) Tropical Forest Biomes
(c) Desert Biomes
(d) Terrestrial Biomes
Answer:
(d) Terrestrial Biomes

Question 5.
The chain of transformation of energy from one group of organisms to another through various trophic levels is called as ……….
(a) Food chain
(b) Food web
(c) Trophic level
(d) None of the above
Answer:
(a) Food chain

Questions 6-8 are Assertion (A) Reason (R) type.
(A) Both Assertion (A) and Reason (R) are true. (R) explains (A).
(B) Both Assertion (A) and Reason (R) are true. (R) does not explain (A).
(C) (A) is true; (R) is false.
(D) Both (A) and (R) are false.

Question 6.
Assertion (A): In Biodiversity each species, no matter how big of small has an important role to play in the ecosystem.
Reason (R): It maintains the ecological balance.
Answer:
(A) Both Assertion (A) and Reason (R) are true. (R) explains (A).

Question 7.
Assertion (A): As the soil is sandy and saline deserts remain agriculturally unproductive. Reason (R): Tribal people who live in deserts practice food gathering and hunting.
Answer:
(B) Both Assertion (A) and Reason (R) are true. (R) does not explain (A).

Question 8.
Assertion (A): Aquatic animals use 10% energy to obtain oxygen.
Reason (R): Terrestrial animals use only 2-4% of energy to obtain oxygen.
Answer:
(D) Both (A) and (R) are false.

III. Answer the following in brief.

Question 1.
What does an ecosystem consists of?
Answer:
An ecosystem consists of three basic components, namely
(A) Abiotic components
(B) Biotic components and
(C) Energy component.

Question 2.
Define Abiotic Components.
Answer:
Abiotic components include the non-living, in-organic, physical and chemical factors in the environment, e.g., Land, Air, Water, Calcium, Iron, etc.

Question 3.
What are energy components?
Answer:
All organisms in the biosphere use energy to work and convert one form of energy into another. The Sun is the ultimate source of energy for the biosphere as a whole. The solar energy gets transformed into other forms of energy through the various components in the ecosystem. The producers, consumers and the decomposers contribute a lot to the energy flow in an ecosystem.

Question 4.
State the unique weather condition that favours thick vegetative cover in the Tropical Forest Biomes.
Answer:
The climate in these biomes shows little seasonal variation with high annual rainfall and relatively constant, high temperature. This unique weather condition favours thick vegetative cover.

Question 5.
How do the people live Desert Biomes?
Answer:
People who live here practice food gathering and hunting. They move their temporary settlements frequently in search of pastures. Transportation becomes very difficult here and is carried on by camels. Reptiles like snakes, lizards, scorpions etc., are most commonly found here.

Question 6.
Mention the Temperate Grasslands in different parts of the world.
Answer:
Temperate grasslands are called differently in different parts of the world.
Prairies – North America
Steppes – Eurasia
Pampas – Argentina and Uruguay
Veld – South Africa
Downs – Australia and Newzealand.

Question 7.
What is an oasis?
Answer:
An oasis is a fertile fresh water source found in deserts and semi-arid regions. Oases are fed by springs. Crops like date palms, figs, citrus fruits, maize etc. are cultivated near these oases.

Question 8.
What is a Biosphere Reserve?
Answer:
A Biosphere Reserve is a special ecosystem or specialized environment with flora and fauna that require protection and nurturing. There are 18 Bioshpere Reserves in India.

IV. Distinguish the following.

Question 1.
Producers Consumers
Answer:

Producers	Consumers
Producers are self nourishing components of the ecosystem. Hence they are called Autotrophs. They are found both on land and water, e.g. Plants, Algae, Bacteria etc.	Consumers are those that depend on producers, directly or indirectly. Hence they are called Heterotrophs.

Question 2.
Abiotic components Biotic components
Answer:

Abiotic components	Biotic Components
Abiotic components include the non-living, inorganic, physical and chemical factors in the environment, e g. Land, Air, Water, Calcium, Iron etc.	Biotic components include plants, animals and micro organisms. Biotic components can be classified into three categories.

Question 3.
Mention any 3 points between Aquatic ecosystem and Terrestrial ecosystem.
Answer:

S.No.	Aquatic Ecosystem	Terrestrial Ecosystem
(i)	Aquatic ecosystem exists on water covering 71% of the earth surface.	Terrestrial ecosystem exists on land covering 29% of the earth surface.
(ii)	Aquatic animals use 20% of energy to obtain oxygen.	Terrestrial animals use only 1-2% of energy to obtain oxygen.
(iii)	In this ecosystem there is abundant of water with limited oxygen supply.	In this ecosystem there is less availability of water, greater availability of gases and temperature fluctuation.

V. Give reasons for the following.

Question 1.
Stable biosphere has to be conserved.
Answer:
A healthy eco system provides clean water, pure water, enriched soil, food, raw materials, medicines etc. Hence stable biosphere has to be conserved.

Question 2.
The Primary cause of today’s loss of biodiversity.
Answer:
The primary cause of today’s loss of biodiversity is habitat alteration caused by human activities. The ever increasing population results in over exploitation of biological resources. This has an adverse impact, on flora and fauna on earth. There are places on earth that are both biologically rich and deeply threatened.

Question 3.
The Net Primary productivity is low in Tundra.
Answer:
This biome experiences long severe winter and short cool summer. Due to the prevailing of low temperature and short growing seasons, the net primary productivity is very low in tundra.

VI. Answer in a paragraph.

Question 1.
Write down the differences between the Aquatic Ecosystem and Terrestrial Ecosystem.
Answer:

S.No.	Aquatic Ecosystem	Terrestrial Ecosystem
(i)	Aquatic ecosystem exists on water covering 71 % of the earth surface.	Terrestrial ecosystem exists on land covering 29% of the earth surface.
(ii)	Aquatic animals use 20% of energy to obtain oxygen.	Terrestrial animals use only 1-2% of energy to obtain oxygen.
(iii)	In this ecosystem there is abundant of Water with limited oxygen supply.	In this ecosystem there is less availability of water, greater availability of gases and temperature fluctuation.
(iv)	The small drifting photosynthetic organisms of the ocean called photo phytoplankton are regarded as the major primary producer.	The primary producer is the plant that produce food through photosynthetic process.
(v)	Aquatic environment is more stable with smaller fluctuation in temperature and other variable.	Terrestrial environment is quite unstable as the land surface is affected by great risks from external impacts.

Question 2.
Desert Biomes.
Answer:
Deserts are usually found on the western margins of the continents between 20° and 30° N and S latitudes. The annual rainfall is less than 25 cm in these regions. Due to the lack of rainfall and arid conditions, these regions do not possess any vegetation but have special vegetation type called Xerophytes. As the soil is sandy and saline, deserts remain agriculturally unproductive. Drought resistant thorny scrubs and bushes, palms are found here.

Tribal people who live here practice food gathering and hunting. They move their temporary settlements frequently in search of pastures. Transportation becomes very difficult here and is carried on by camels. Reptiles like snakes, lizards, scorpions etc., are most commonly found here.

Students can Download Tamil Chapter 3.3 தேசியம் காத்த செம்மல் பசும்பொன் உ.முத்துராமலிங்கத்தேவர் Questions and Answers, Summary, Notes Pdf, Samacheer Kalvi 7th Tamil Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 1 Chapter 3.3 தேசியம் காத்த செம்மல் பசும்பொன் உ.முத்துராமலிங்கத்தேவர்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
முத்துராமலிங்கத்தேவர் முதன்முதலில் உரையாற்றிய இடம் …….
அ) தூத்துக்குடி
ஆ) காரைக்குடி
இ) சாயல்குடி
ஈ) மன்னார்குடி
Answer:
இ) சாயல்குடி

Question 2.
முத்துராமலிங்கத்தேவர் நடத்திய இதழின் பெயர்…
அ) இராஜாஜி
ஆ) நேதாஜி
இ) காந்திஜி
ஈ) நேருஜி
Answer:
ஆ) நேதாஜி

Question 3.
தேசியம் காத்த செம்மல் எனப் பசும்பொன் முத்துராமலிங்கத்தேவரைப் பாராட்டியவர்…….
அ) இராஜாஜி
ஆ) பெரியார்
இ) திரு.வி.க.
ஈ) நேதாஜி
Answer:
இ) திரு.வி.க.

குறுவின

Question 1.
முத்துராமலிங்கத்தேவரைப் பாராட்டிப் பெரியார் கூறியது யாது?
Answer:
(i) தேசியம் உடல், தெய்வீகம் உயிர் எனக் கருதி மக்கள் தொண்டு செய்தவர் பசும்பொன் முத்துராமலிங்கத் தேவர்.

(ii) இவர் வீரப்பேச்சால் எத்தனையோ தியாகிகளையும் விவேகப் பேச்சால் எத்தனையோ அறிவாளிகளையும் உண்டாக்கியவர்; உண்மையை மறைக்காமல் வெளியிடுவதில்  தனித் துணிச்சல் பெற்றவர்; சுத்தத் தியாகி என்று தந்தைப் பெரியார் பாராட்டினார்.

Question 2.
முத்துராமலிங்கத்தேவரின் பேச்சுக்கு வாய்ப்பூட்டுச் சட்டத்தின் மூலம் தடைவிதிக்கப்படக்  காரணம் யாது?
Answer:

• முத்துராமலிங்கத்தேவர் மேடைகளில் ஆற்றிய வீர உரையைக் கேட்ட மக்கள் ஆங்கில ஆட்சிக்கு எதிராக வீறுகொண்டு எழுந்தனர்.
• அதனால் அச்சமடைந்த ஆங்கில அரசு பலமுறை அவரைக் கைது செய்து சிறையில் அடைத்தது. மேலும் வாய்ப்பூட்டுச் சட்டம் மூலம் மேடைகளில் அரசியல் பேசக்கூடாது என்று அவருக்குத் தடை விதித்தது.

Question 3.
முத்துராமலிங்கத்தேவர் பெற்றிருந்த பல்துறை ஆற்றலைப் பற்றி எழுதுக.
Answer:
முத்துராமலிங்கத்தேவர் பெற்றிருந்த பல்துறை ஆற்றல் : சிலம்பம், குதிரை ஏற்றம், துப்பாக்கிச் சுடுதல், சோதிடம், மருத்துவம் போன்ற பலதுறைகளிலும் ஆற்றல் உடையவராக விளங்கினார்.

சிறுவினா

Question 1.
நேதாஜியுடன் முத்துராமலிங்கத்தேவர் கொண்ட தொடர்புப் பற்றி எழுதுக.
Answer:
(i) வங்கச் சிங்கம் என்று போற்றப்பட்ட நேதாஜி சுபாஷ் சந்திரபோசுடன் நெருங்கிய தொடர்பு கொண்டிருந்தார். அவரைத் தமது அரசியல் குருவாக ஏற்றுக் கொண்டார்.

(ii) முத்துராமலிங்கத்தேவரின் அழைப்பை ஏற்றுக் கி.பி. (பொ.ஆ.) 1939-ஆம் ஆண்டு செப்டம்பர்த் திங்கள் ஆறாம் நாள் நேதாஜி மதுரைக்கு வருகை தந்தார்.

(iii) நேதாஜி தொடங்கிய இந்திய தேசிய இராணுவத்தில் முத்துராமலிங்கத்தேவரின் முயற்சியால் ஏராளமான தமிழர்கள் இணைந்தனர். விடுதலைக்குப் பின் நேதாஜி என்னும் பெயரில் வார இதழ் ஒன்றையும் நடத்தினார்.

Question 2.
தொழிலாளர் நலனுக்காக முத்துராமலிங்கத்தேவர் செய்த தொண்டுகள் யாவை?
Answer:
(i) 1938 காலக்கட்டத்தில் மதுரையில் 23 தொழிலாளர் சங்கங்களின் தலைவராகத் தேவர் திகழ்ந்தார்.

(ii) மதுரையிலிருந்த நூற்பு ஆலை ஒன்றில் வேலை செய்த தொழிலாளர்களின்  உரிமைக்காகத் தோழர் ப. ஜீவானந்தத்துடன் இணைந்து 1938 ஆம் ஆண்டு போராட்டம் நடத்தினார்.

(iii) அதற்காக ஏழு திங்கள் சிறைத் தண்டனைப் பெற்றார். உழவர்களின் நலன் காக்க இராஜபாளையத்தில் மிகப்பெரிய அளவிலான மாநாடு ஒன்றை நடத்தினார். பெண் தொழிலாளர்களுக்கு மகப்பேறு காலத்தில் ஊதியத்துடன் கூடிய விடுப்பு வேண்டும் என்று போராடினார்.

சிந்தனை வினா

Question 1.
சிறந்த தலைவருக்குரிய பண்புகள் எவை என நீங்கள் கருதுகிறீர்கள்?
Answer:
சிறந்த தலைவருக்குரிய பண்புகள் :
(i) நல்ல குண இயல்புகள் கொண்டு இருக்க வேண்டும். செயல்பாடுகள் நல்லதாக இருக்க வேண்டும். வெறும் அறிவு மட்டுமே இருந்தால் ஒருவருக்கு தலைமைப்பண்பு இருக்கின்றது என்ற கருத முடியாது. அறிவு, அனுபவம், மனிதர்களை மதித்து நடந்து கொள்ளும் பண்பு ஆகியவற்றைக் கொண்டவரே சிறந்த தலைவர்.

(ii) மக்களின் வறுமைகளையும், இன்னல்களையும் போக்குபவரே சிறந்தத் தலைவர். கீழ்மட்டத்திலுள்ள மக்களுக்கு உரிமைகளைப் பெற்றுத் தர வேண்டும்; மேல் மட்டத்திலுள்ள மக்களின் நலனையும் பேணி காக்க வேண்டும்.

(iii) இருவரையும் சமமாக நடத்த வேண்டும். உயர்ந்தவர் தாழ்ந்தவர் எனப் பிரிக்காமல், இருவரும் ஒருவரே என அரவணைத்து செல்பவரே சிறந்த தலைவருக்குரிய பண்புகளில் ஒன்றாகும்.

(iv) சிறந்த தலைவரானவர் பிறரது மனதில் தாக்கத்தை ஏற்படுத்தி திறமையாகச் செயல்பட வைக்கும் ஆற்றல் உள்ளவர்களாக இருப்பர். இதுவரை வெளிப்படுத்தாமலிருந்த திறமைகளையும் மக்களிடமிருந்து வெளிக்கொணரும் வகையில் இவர்களது செயல்பாடு அமையும், இவர்களது அணுகுமுறையும், செயல்பாடுகளும் பிறரையும் சிறப்பாக செயல்படத் தூண்டும் வகையில் இருக்கும்.

(v) மக்களிடத்தில் எந்தவித பாகுபாடும் இல்லாமல், பய உணர்ச்சியும் இல்லாமல், நடந்து கொள்ளுதல், செயல்களுக்கு பொறுப்பேற்றுக் கொண்டு செயல்படுதல், வெற்றி என்றாலும் சரி, தோல்வி என்றாலும் சரி அதற்கு பொறுப்பேற்றுக் கொள்ளும் மனப் பக்குவம் போன்ற தலைமைப் பண்புகள் கொண்டவர்களே! சிறந்த தலைவராக முடியும்.

கற்பவை கற்றபின்

Question 1.
நாட்டுக்கு உழைத்த சிறந்த பிற தலைவர்களைப் பற்றிய செய்திகளைத் திரட்டி எழுதுக.
Answer:
சுபாஷ் சந்திரபோஸ்(நேதாஜி) :

வங்காளத்தில் 1897-ஆம் ஆண்டு ஜனவரி 23-ஆம் நாள் கோதாலியா என்ற சிற்றூரில் ஜானகிநாத் – பிரபாவதி தம்பதிகளுக்கு மகனாகப் பிறந்தவர் சுபாஷ் சந்திரபோஸ். தாயார் பிரபாவதியிடம் வீரம் நிறைந்த காளிதேவியின் கதைகளைக் கேட்டு அறிவார். 1915 ஆம் ஆண்டு பள்ளியின் மாணவத் தலைவனாக தேர்வு செய்யப்பட்டார். ஐரோப்பாவில் வாழ்ந்த வெளிநாட்டுவாழ் இந்தியர்கள் சுபாஷை “நேதாஜி” என்று அழைத்தனர். அன்று முதல் சுபாஷ் சந்திரபோஸ் “நேதாஜி” என்று அழைக்கப்படுகிறார்.

இந்து, முஸ்லீம் என்று பிரிந்து கிடந்த இந்தியர்களை ஒன்றிணைக்க நேதாஜியால் அப்போது “ஜெய்ஹிந்த்” என்ற தாரக மந்திரம் எழுப்பப்பட்டது.

1945-ஆம் ஆண்டு ஆகஸ்டு 16-ஆம் நாள் ஜப்பானுடன் பேச்சுவார்த்தை நடத்த சைகோனிலிருந்து விமானத்தில் நேதாஜி புறப்பட்டார். ஆனால் வழியில் (தைவான்) தைஹோக் என்ற இடத்தில் அவர் சென்ற விமானம் விபத்துக்குள்ளானது என்றும், இதில் பலத்த காயமடைந்த நேதாஜி ஆகஸ்டு 18-ஆம் நாள் தனது இன்னுயிர் நீத்தார் என்றும் கூறப்பட்டது.

அம்பேத்கர் :

மகாராஷ்டிரா மாநிலம் ரத்னகிரி மாவட்டத்தில் அம்பாலாதே கிராமத்தில் 1891-ஆம் ஆண்டு ஏப்ரல் 14-ஆம் நாள் அம்பேத்கர் பிறந்தார். பெற்றோர் சுபேதார் இராம்ஜி சக்பால் – பீமாபாய் ஆவர். பெற்றோர் இட்ட பெயர் பீமாராவ் இராம்ஜி ஆகும்.

1913-ஆம் ஆண்டு கொலம்பியா பல்கலைக்கழத்தில் அரசியல் மற்றும் பொருளாதாரம் ஆகிய பிரிவுகளில் மேல் படிப்புக்காகச் சேர்ந்தார். 1915-ஆம் ஆண்டு எம்.ஏ. பட்டமும், 1916ஆம் ஆண்டு தத்துவப் பேரறிஞர் பட்டமும் பெற்றார்.

இந்திய நாட்டுக்குத் திரும்பியதும் பம்பாய் சைடன்ஹாம் கல்லூரியில் 1918 ஆம் ஆண்டிலிருந்து 1920-ஆம் ஆண்டு வரை பொருளாதாரப் பேராசிரியராகப் பணியாற்றினார்.

மீண்டும் லண்டன் சென்று 1923-இல் எம்.எஸ்.சி மற்றும் பாரிஸ்டர் பட்டங்களைப் பெற்றார். பம்பாயில் வழக்கறிஞர் தொழில் தொடங்கினார். 1956ஆம் ஆண்டு அம்பேத்கர் புத்த மதத்தில் இணைந்தார். அதே ஆண்டில் டிசம்பர் 6-ஆம் நாள் இந்திய அரசு 1990 ஆம் ஆண்டு பாரத ரத்னா” விருது வழங்கி கௌரவித்தது.

கூடுதல் வினாக்கள்

நிரப்புக :

Question 1.
சன்மார்க்க சண்டமாருதம் என்று அழைக்கப்படுபவர் ……
Answer:
முத்துராமலிங்கத்தேவர்

Question 2.
பசும்பொன் முத்துராமலிங்கத் தேவரை ‘தென்னாட்டுச் சிங்கம்’ எனக் கூறியவர்
Answer:
அறிஞர் அண்ணா

Question 3.
முத்துராமலிங்கத் தேவர் மதுரையிலிருந்த நூற்பு ஆலை ஒன்றில் வேலைசெய்த தொழிலாளர்களின் உரிமைக்காக …………… உடன் இணைந்து 1938 ஆம் ஆண்டு போராட்டம் நடத்தினார்.
Answer:
ப.ஜீவானந்தம்

விடையளி :

Question 1.
பசும்பொன் முத்துராமலிங்கத்தேவர் சிறை வாசம் பற்றி எழுது.
Answer:
பசும்பொன் முத்துராமலிங்கத்தேவர் சுதந்திரப் போராட்டத்தில் மிகத் தீவிரமாக ஈடுபட்டதால் கைது செய்யப்பட்டு அலிப்பூர், அமராவதி, தாமோ, கல்கத்தா, சென்னை, வேலூர் போன்ற சிறைகளில் சிறை வைக்கப்பட்டிருந்தார். இரண்டாம் உலகப்போர் சமயத்தில் மத்திய பிரதேசத்தின் தாமோ என்னும் நகரில் உள்ள இராணுவச் சிறையில் அடைக்கப்பட்டுப் போர் முடிந்த பிறகுதான் விடுதலை செய்யப்பட்டார்.

Question 2.
முத்துராமலிங்கத்தேவரின் சிறப்புப் பெயர்கள் யாவை?
Answer:
தேசியம் காத்த செம்மல், வித்யா பாஸ்கர், பிரவசன கேசரி, சன்மார்க்க சண்டமாருதம், இந்து புத்த சமயமேதை.

Students can Download Tamil Chapter 3.5 வழக்கு Questions and Answers, Summary, Notes Pdf, Samacheer Kalvi 7th Tamil Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 1 Chapter 3.5 வழக்கு

மதிப்பீடு

1. பந்தர் – முதற்போலி
2. மைஞ்சு – முற்றுப்போலி
3. அஞ்சு – இடைப்போலி
4. அரையர் – கடைப்போலி
Answer:
1. பந்தர் – கடைப்போலி
2. மைஞ்சு – முதற்போலி
3. அஞ்சு – முற்றுப்போலி
4. அரையர் – இடைப்போலி

குறுவினா

Question 1.
வழக்கு என்றால் என்ன?
Answer:
நம் முன்னோர் எந்தெந்தச் சொற்களை என்னென்ன பொருளில் பயன்படுத்தினார்களோ அச்சொற்களை அவ்வாறே பயன்படுத்துவதை வழக்கு என்பர்.

Question 2.
தகுதி வழக்கின் வகைகள் யாவை?
Answer:
தகுதி வழக்கின் வகைகள்.
தகுதி வழக்கு மூன்று வகைப்படும். அவை

• இடக்கரடக்கல்
• மங்க லம்
• குழூஉக்குறி

Question 3.
வாழைப்பழம் மிகவும் நஞ்சு விட்டது – இத்தொடரில் இடம்பெற்றுள்ள போலிச் சொல்லைக் கண்டறிக. அதன் சரியான சொல்லை எழுதுக.
Answer:
இத்தொடரில் இடம் பெற்றுள்ள போலிச் சொல் நஞ்சுவிட்டது.
அதன் சரியான சொல் ‘நைந்து’

கற்பவை கற்றபின்

Question 1.
மூவகைப் போலிகளிலும் எந்தெந்த எழுத்துகள் எந்தெந்த எழுத்துகளாக மாறுகின்றன என்பதை அறிந்து தொகுக்க.
Answer:
முதற்போலி :

அகரம் ஐகாரத்தை ஒத்திருத்தல் :
(i) பசல் – பைசல்
(ii) வயிரம் – வைரம்
(iii) மஞ்சு – மைஞ்சு
(iv) மயல் – மையல்

நகரத்திற்கு ஞகரம் :

(i) நயம் – ஞயம்
(ii) நான் – ஞான்

மகரத்திற்கு நகரம் :

(i) முப்பது – நுப்பது
(ii) முனை – நுனை

இடைப்போலி :

(i) அமச்சன் – அமைச்சன்(அ – ஐ)
(ii) அரயன் – அரையன் (அ – ஐ)
(iii) நேசம் – நேயம் (ச-ய)
(iv) என்ப ர் – என்ம ர் (ப-ம் )
(v) ஐந்நூறு – ஐஞ்நூறு (ந -ஞ)
(vi) அசல் – அயல் (ச – ய)

கடைப்போலி :

முற்றுப்போலி :

இலக்கணப்போலி : முன்பின் தொக்கப் போலி

கூடுதல் வினாக்கள்

நீரப்புக.

Question 1.
எழுத்திலும் பேச்சிலும் சொற்களைப் பயன்படுத்தும் முறை………….. எனப்படும்.
Answer:
வழக்கு

Question 2.
…………… என்னும் சொல்லைப் பேச்சு வழக்கில் வாசல் என வழங்குகிறோம்.
Answer:
வாயில்

Question 3.
இலக்கணப்போலி என்பது பெரும்பாலும் சொற்களின் …………… இடம் மாறி வருவதையே குறிக்கும்.
Answer:
முன்பின் பகுதிகள்

Question 4.
நாகரிகம் கருதி மறைமுகமாகக் குறிப்பிடுதல் ……………
Answer:
இடக்கரடக்கல்

Question 5.
மங்கலமற்ற சொற்களை மாற்றி மங்கலச் சொற்களால் குறிப்பிடுதல் ………..
Answer:
மங்கலம்

விடையளி :

Question 1.
இயல்பு வழக்கு என்றால் என்ன? அவை எத்தனை வகைப்படும்?
Answer:
ஒரு பொருளை அதற்கே உரிய இயல்பான சொற்களால் குறிப்பிடுவது இயல்பு வழக்கு ஆகும். இயல்பு வழக்கு மூன்று வகைப்படும்.

1. இலக்கணமுடையது
2. இலக்கணப்போலி
3. மரூஉ .

Question 2.
இலக்கணமுடையது பற்றி எழுதுக.
Answer:
நிலம், மரம், வான், எழுது – ஆகிய சொற்கள், தமக்குரிய பொருளை எவ்வகை மாறுபாடும் இல்லாமல் இயல்பாகத் தருகின்றன. இவ்வாறு இலக்கண நெறி மாறாமல் முறையாக அமைந்த சொல் இலக்கணமுடையது ஆகும்.

Question 3.
இலக்கணப் போலி சான்றுடன் விளக்குக.
Answer:
இல்லத்தின் முன் பகுதியை இல்முன் எனக் குறிக்க வேண்டும். ஆனால் அதனை நம் முன்னோர் முன்றில் என மாற்றி வழங்கினர். கிளையின் நுனியைக் கிளை நுனி எனக் கூறாமல் நுனிக்கிளை எனக் குறிப்பிடுகிறோம். இவ்வாறு இலக்கண முறைப்படி அமையாவிடினும், இலக்கணமுடையவை போலவே ஏற்றுக் கொள்ளப்படும் சொற்கள் இலக்கணப்போலி எனப்படும்.

இலக்கணப்போலி என்பது பெரும்பாலும் சொற்களின் முன்பின் பகுதிகள் இடம்மாறி வருவதையே குறிக்கும். எனவே, இலக்கணப் போலியை முன்பின்னாகத் தொக்க போலி எனவும் குறிப்பிடுவர்.

எ.கா. புறநகர், கால்வாய், தசை, கடைக்கண்.

Question 4.
மரூஉ என்றால் என்ன?
Answer:
நாம் எல்லாச் சொற்களையும் எல்லா இடங்களிலும் முழுமையான வடிவத்தில் பயன்படுத்துவது இல்லை. தஞ்சாவூர் என்னும் பெயரைத் தஞ்சை என்றும், திருநெல்வேலி என்னும் பெயரை நெல்லை எனவும் வழங்குகிறோம். இவ்வாறு இலக்கண நெறியிலிருந்து பிறழ்ந்து, சிதைந்து வழங்கும் சொற்கள் மரூஉ எனப்படும்.
எ.கா. கோவை, குடந்தை, எந்தை, போது, சோணாடு.

Question 5.
தகுதி வழக்கு என்றால் என்ன? அவை எத்தனை வகைப்படும்? அதன் வகைகளை விளக்குக.
Answer:

1. ஏதேனும் ஒரு காரணத்தினால் பிறரிடம் சொல்லத் தகுதியற்ற சொற்களைத் தகுதியான வேறு சொற்களால் குறிப்பிடுவது தகுதி வழக்கு ஆகும்.
2. தகுதி வழக்கு மூன்று வகைப்படும். அவை
3. இடக்கரடக்கல், மங்கலம், குழூஉக்குறி

இடக்கரடக்கல் :

பிறரிடம் வெளிப்படையாகச் சொல்லத்தகாத சொற்களைத் தகுதியுடைய வேறு சொற்களால் கூறுவது இடக்கரடக்கல் ஆகும்.
எ.கா. : கால் கழுவி வந்தான், குழந்தை வெளியே போய்விட்டது. ஒன்றுக்குப் போய் வந்தேன்.

மங்க லம் :
செத்தார் என்பது மங்கலமில்லாத சொல் என நம் முன்னோர் கருதினர். எனவே, செத்தார் எனக் குறிப்பிடாமல் துஞ்சினார் எனக் குறிப்பிட்டனர். நாம் இக்காலத்தில் இயற்கை எய்தினார் என்று குறிப்பிடுகிறோம். இவ்வாறு மங்கலமில்லாத சொற்களை மங்கலமான வேறு சொற்களால் குறிப்பதை மங்கலம் என்பர்.

எ.கா. : ஓலை – திருமுகம்
கறுப்பு ஆடு – வெள்ளாடு
விளக்கை அணை – விளக்கைக் குளிரவை
சுடுகாடு – நன்காடு

குழூஉக்குறி :

பலர் கூடியிருக்கும் இடத்தில் சிலர் மட்டும் தமக்குள் சில செய்திகளைப் பகிர்ந்து கொள்ள விரும்பினால் மற்றவர்கள் புரிந்து கொள்ள இயலாத வகையில் சொற்களைப் பயன்படுத்துவர். இவ்வாறு ஒரு குழுவினார் ஒரு பொருள் அல்லது செயலைக் குறிக்கத் தமக்குள் பயன்படுத்திக் கொள்ளும் சொற்கள் குழூஉக்குறி எனப்படும்.

எ.கா. : பொன்னைப் பறி எனல் (பொற்கொல்லர் பயன்படுத்துவது)
ஆடையைக் காரை எனல் (யானைப்பாகர் பயன்படுத்துவது)

Question 6.
போலி என்றால் என்ன?
Answer:
சொல்லின் முதலிலோ, இடையிலோ, இறுதியிலோ இயல்பாக இருக்க வேண்டிய ஓர் எழுத்திற்குப் பதிலாக வேறு ஓர் எழுத்து இடம்பெற்று அதே பொருள் தருவது போலி எனப்படும். போலி என்னும் சொல் போல இருத்தல்’ என்பதிலிருந்து தோன்றியது. போலி மூன்று வகைப்படும்.

Question 7.
முதற்போலி என்றால் என்ன?
Answer:
பசல் – பைசல், மஞ்சு – மைஞ்சு, மயல் – மையல் ஆகிய சொற்களில் முதல் எழுத்து மாறினாலும் பொருள் மாறவில்லை. இவ்வாறு சொல்லின் முதலில் இருக்க வேண்டிய எழுத்திற்குப் பதிலாக வேறு ஓர் எழுத்து அமைந்து அதே பொருள் தருவது முதற்போலியாகும்.

Question 8.
இடைப்போலி என்றால் என்ன?
Answer:
சொல்லின் இடையில் இருக்க வேண்டிய எழுத்திற்குப் பதிலாக வேறு ஓர் எழுத்து அமைந்து அதே பொருள் தருவது இடைப்போலியாகும்.
(எ.கா) அமச்சு – அமைச்சு
இலஞ்சி – இலைஞ்சி
அரயர் – அரையர்

Question 9.
கடைப்போலி என்றால் என்ன?
Answer:
சொல்லின் இறுதியில் இருக்க வேண்டிய எழுத்திற்குப் பதிலாக வேறு ஓர் எழுத்து அமைந்து அதே பொருள் தருவது கடைப்போலியாகும். அஃறிணைப் பெயர்களின் இறுதியில் நிற்கும் மகர எழுத்திற்குப் பதிலாக னகரம் கடைப்போலியாக வரும். லகர எழுத்திற்குப் பதிலாக ரகரம் கடைப்போலியாக வரும்.

(எ.கா) அகம் – அறன்
நிலம் – நிலன்
முகம் – முகன்
பந்தல் – பந்தர்
சாம்பல் – சாம்பர்

Question 10.
முற்றுப்போலி என்றால் என்ன?
Answer:
ஒரு சொல்லில் இயல்பாக அமைந்த எழுத்துகளுக்குப் பதிலாக எழுத்துகள் அனைத்தும் வேறுபட்டாலும் பொருள் மாறாமல் இருப்பது முற்றுப்போலி எனப்படும்.
(எ.கா) ஐந்து – அஞ்சு

மொழியை ஆள்வோம்

கேட்க.

Question 1.
பசும்பொன் முத்துராமலிங்கத்தேவர் பேச்சின் ஒலிப்பதிவைக் கேட்டு மகிழ்க.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டியவை.

கீழ்க்காணும் தலைப்புகளுள் ஒன்று பற்றி இரண்டு நிமிடம் பேசுக

Question 1.
தேசியம் காத்த செம்மல்
Answer:
முன்னுரை :
தேசியம் காத்த செம்மல், வேதாந்த பாஸ்கர், பிரணவகேசரி, சன்மார்க்க சண்டமாருதம், இந்து புத்த சமய மேதை எனப் பலவாறு பாராட்டப்பட்டவர் முத்துராமலிங்கத் தேவர்.

பிறப்பு மற்றும் பெற்றோர் :
உக்கிரபாண்டித் தேவருக்கும், இந்திராணி அம்மையாருக்கும் 1908 ஆம் ஆண்டு அக்டோபர் மாதம் முப்பதாம் நாள் பசும்பொன் என்ற ஊரில் இவர் பிறந்தார். பிள்ளைப் பருவத்திலேயே தாயை இழந்ததால் இவருக்கு இஸ்லாமியப் பெண்மணி ஒருவர் தாயாகப் பாலூட்டி வளர்த்தார். பாட்டியின் வளர்ப்பில் வளர்ந்தவர்.

கல்வி :
தேவர், இராமநாதபுரம் அரசு உயர்நிலைப் பள்ளியில் பத்தாம் வகுப்புப் படித்தார். அங்கு பிளேக் நோய் பரவியதால் கல்வி தடைப்பட்டது. எனினும் கேள்வி அறிவும் பட்டறிவும் பெற்று, ஆங்கிலத்திலும் தமிழிலும் மேடைகளில் பேசும் வல்லமை பெற்றார். எடுத்துக்காட்டாக, ஒருமுறை காசி இந்துப் பல்கலைக்கழகத்தில் அவர் ஆங்கிலத்தில் உரையாற்றியதைக் குறிப்பிடலாம். அனைவரும் அவருடைய பேச்சில் மயங்கினர். அக்கூட்டத்திற்குத் தலைமை தாங்கிய சர்.சி.பி. இராமசாமி ஆங்கிலம் உலகை ஆள்கிறது. நம் தேவர் மூன்று மணிநேரம் ஆங்கிலத்தை அடக்கி ஆண்டார்’ என்று பாராட்டினார்.

அரசியல் : ‘
அரசியல் வாழ்வில் தேவர், மேன்மை பெற்று விளங்கினார். தமிழக்கத்தில் நடைபெற்ற தேர்தல்களில் தொடர்ந்து ஐந்து முறை வெற்ற வெற்றிகள் மக்களிடம் அவருக்கிருந்த செல்வாக்கைக் காட்டின. அவர் எப்போதும் ஓட்டு கேட்பதற்காகத் தொகுதிகளுக்குச் சென்றதில்லை. தொண்டு செய்வதற்காக மட்டுமே செல்வார்.

தேவர் வாழ்ந்த விதம் :
தேவர் விவேகானந்தரின் தூதராவும், நேதாஜியின் தளபதியாகவும், சத்திய சீலராகவும், முருக பக்தராகவும், ஆன்மீகப் புத்திரராகவும், தமிழ் பாடும் சித்தராகவும், தென்பாண்டிச் சீமையின் முடிசூடா மன்னராகவும், புலமையில் கபிலராகவும், இந்தியத் தாயின் நன்மகனாகவும் தேசியம் காத்த செம்மலாக வாழ்ந்தார்.

மனித குலத்தின் வழிகாட்டி :
மனித மனநிலையை இருள், மருள், தெருள், அருள் எனக் குறிப்பிட்டார். இறப்பு என்பது எல்லா வகைகளிலும் வரலாம். அதாவது, பனை மரத்திலிருந்து விழுந்தவன் பிழைத்ததும் உண்டு. வயல் வரப்பில் வழுக்கி விழுந்தவன் இறந்ததும் உண்டு என்று கூறுவார். இவ்வகையில் சமயம், சமுதாயம் குறித்த இவருடைய சிந்தனைகள், மனித குலத்திற்கு வழிகாட்டுவனவாயின.

முடிவுரை :
தெய்வத் திருமகனார் தேவர் அவர்கள் தம் பேச்சாற்றலால் இலக்கியப் பணி, சமுதாயப் பணி, அரசியல் பணி என அனைத்தையும் ஒரு சேரச் செய்தவர். 1963 ஆம் ஆண்டு அக்டோபர் மாதம் 30 ஆம் இவ்வுலகை விட்டு இவருடைய இன்னுயிர் பிரிந்தது. அவருடைய பொன்னுடல் மறைந்தாலும் புகழுடல் மறையவில்லை.

Question 2.
கப்பலோட்டிய தமிழர்
Answer:
முன்னுரை :
‘சிதம்பரனாரின் பிரசங்கத்தைக் கேட்டால் செத்த பிணம் உயிர்பெற்று எழும், புரட்சி ஓங்கும் அடிமைப்பட்ட நாடு ஐந்தே நிமிடங்களில் விடுதலை பெறும்’ என்ற கூற்றின் படி பேச்சாற்றல் மிக்க வ. உ. சிதம்பரனாரின் வாழ்வியல் பற்றி இக்கட்டுரையில் காண்போம்.

இளமை காலம் :
சிதம்பரனார் 1872 ஆம் ஆண்டு செப்டம்பர் 5 ஆம் நாள் தமிழ்நாட்டில் தூத்துக்குடி மாவட்டம் வண்டானம் ஒட்டப்பிடாரம் என்ற ஊரில் பிறந்தார். பெற்றோர் உலகநாத பிள்ளை , பரமாயி அம்மாள் ஆவர்.

கல்லூரிப் படிப்பை முடித்த பின்னர் தன் தந்தையைப் போன்று வழக்கறிஞரானார். வ.உ.சி. ஏழைகளுக்காக வாதாடினார். சில சமயங்களில் கட்டணம் பெறாமலும் வாதாடினார். சிறந்த வழக்கறிஞர் என்று போற்றப்பட்டார்.

வெள்ளையர்களின் வீழ்ச்சி :
சுதேசக் கப்பலின் வருகையால் வெள்ளையர்களின் கப்பல் வாணிகம் தளர்ந்தது. வெள்ளையர்கள் பல சூழ்ச்சிகளைச் செய்தனர். வ.உ.சிதம்பரனாருக்குக் கையடக்கம் தருவதாகவும் கூறினர். பலரைப் பயமுறுத்தினர். இறுதியில் அடக்குமுறையைக் கையாண்டனர்.

சிறையில் தமிழ்ப்பணி :
சிறைச்சாலையில் செக்கிழுத்த துயரத்தையும், கடும் பணிபுரிந்தபோது வந்த கண்ணீரையும் தமிழ் நூல்களைப் படித்து மாற்றிக்கொண்டார். தொல்காப்பியம், இன்னிலை ஆகிய நூல்களைப் படித்தார். ஆலன் என்பவர் இயற்றிய ஆங்கில மொழி நூலை மனம் போல் வாழ்வு’ என்று தமிழில் மொழிபெயர்த்தார். மெய்யறிவு, மெய்யறம் என்ற நூல்களை இயற்றினார்.

முடிவுரை :
வ.உ.சிதம்பரனார் சிறையிலிருந்து விடுவிக்கப்பட்டவுடன் துறைமுகத்தை வந்து பார்த்தார். சுதேசக் கப்பலைக் காணாமல் துயருற்றார். பட்ட பாடெல்லாம் பயனற்றுப் போயிற்றே என்று பரிதவித்தார். விடுதலைக்காகப் போராடி நாட்டு மக்களின் துயர் துடைத்த வ.உ.சி. அவர்கள் 1936 ஆம் ஆண்டு நவம்பர் மாதம் 18 ஆம் நாள் இயற்கை எய்தினார்.

சொல்லக் கேட்டு எழுதுக

1. அவன் எங்குள்ளான் என எனக்குத் தெரியவில்லை .
2. வீடுகள் தோறும் மணிகளால் அழகுசெய்யப்பட்ட மேடைகள் இருக்கும்.
3. தெய்வீகத்தையும் தேசியத்தையும் தமது இரு கண்களாகக் கருதியவர்
   முத்துராமலிங்கத்தேவர்.
4. தொல்காப்பியத்தைப் படித்துத் தொல்லையெல்லாம் மறந்தேன்.
5. இலக்கண நெறி மாறாமல் முறையாக அமைந்த சொல் இலக்கணமுடையது ஆகும்.

அறிந்து பயன்படுத்துவோம்

ஒரு தொடரில் மூன்று பகுதிகள் இடம்பெறும்.
அவை 1. எழுவாய், 2. பயனிலை, 3. செயப்படுபொருள்.
ஒரு தொடரில் யார்? எது? எவை? என்னும் வினாக்களுக்கு விடையாக அமைவது எழுவாய்.

எடுத்துக்காட்டு:

• நீலன் பாடத்தைப் படித்தான்.
• பாரி யார்?
• புலி ஒரு விலங்கு. இத்தொடர்களில் நீலன், பாரி, புலி ஆகியன எழுவாய்கள்.

ஒரு தொடரை வினை, வினா, பெயர் ஆகியவற்றுள் ஏதேனும் ஒன்றைக் கொண்டு முடித்து வைப்பது பயனிலை.

எடுத்துக்காட்டு :

• கரிகாலன் கல்லணையைக் கட்டினான்.
• கரிகாலன் யார்?
• கரிகாலன் ஒரு மன்ன ன்.

இத்தொடர்களில் கட்டினான், யார், மன்னன் ஆகியன பயனிலைகள்.
யாரை, எதை, எவற்றை என்னும் வினாக்களுக்கு விடையாக வருவது செயப்படுபொருள்.

எடுத்துக்காட்டு :

• நான் கவிதையைப் படித்தேன்.
• என் புத்தகத்தை எடுத்தது யார்?
• நெல்லிக்கனியைத் தந்தவர் அதியமான்.

இத்தொடர்களில் கவிதை, புத்தகம், நெல்லிக்கனி ஆகியன செயப்படுபொருள்கள்.

பின்வரும் தொடர்களை எழுவாய், பயனிலை, செயப்படுபொருள் எனப் பிரிக்க

1. வீரர்கள் நாட்டைக் காத்தனர்
2. பொதுமக்கள் அந்நியத்துணிகளைத் தீயிட்டு எரித்தனர்.
3. கொற்கைத் துறைமுகத்திலே பாண்டியனுடைய மீன்கொடி பறந்தது.
4. திருக்குறளை எழுதியவர் யார்?
5. கபிலர் குறிஞ்சிப்பாட்டை எழுதிய புலவர்.

Answer:

எழுதிய புலவர் குறிஞ்சிப்பாட்டை எழுவாய், பயனிலை, செயப்படுபொருள் ஆகிய மூன்றும் அமையும்படி ஐந்து தொடர்களை எழுதுக.

Answer:

1. சிறுவன் மிதிவண்டியை ஓட்டினான்.
2. மன்னர் நல்லமுறையில் நாட்டை ஆண்டார்.
3. குழந்தை பழத்தைத் தின்றது.
4. மாணவர்கள் பாடத்தைப் படித்தனர்.
5. ஓவியர் ஓவியத்தை வரைந்தார்.

கீழ்க்காணும் தலைப்பில் கட்டுரை எழுதுக

நான் விரும்பும் தலைவர்

திரு.வி. கல்யாணசுந்தரனார் :

முன்னுரை :

திரு.வி.க. காஞ்சிபுரம் மாவட்டத்தில் உள்ள துள்ளம் என்னும் சிற்றூரில் பிறந்தார். இவருடைய பெற்றோர் விருத்தாசலம், சின்னம்மா. இவரின் முன்னோர்கள் திருவாரூரைச் சேர்ந்தவர்கள்.

இளமைக் காலம் :

தொடக்கத்தில் தம் தந்தையிடம் கல்வி பயின்றார். பின்னர் சென்னை இராயப்பேட்டையில் வெஸ்லி பள்ளியில் பயின்றார். நான்காம் வகுப்பு படிக்கும் போது உடல் நிலை பாதிக்கப்பட்டதால் படிப்பு தடைப்பட்டது.
வெஸ்லி பள்ளியில் ஆசிரியராக இருந்த யாழ்ப்பாணம் நா.

கதிர்வேற்பிள்ளை என்பவரிடம் தமிழும் மயிலை தணிகாசல முதலியாரிடம் தமிழ் மற்றும் சைவ நூல்களையும் கேட்டறிந்தார். பாம்பன் சுவாமிகளிடம் உபநிடதங்களையும் ஜஸ்டிஸ் சதாசிவராவ் அவர்களின் தொடர்பால் ஆங்கில அறிவையும் பெற்றார்.

விடுதலை இயக்கம் :

விடுதலை இயக்கத்தில் ஈடுபாடு ஏற்பட்டது. தேசபக்தன்’ பத்திரிகையின் ஆசிரியராக இரண்டரை ஆண்டுகள் இருந்தார். தனது எழுச்சிமிக்க எழுத்துகளால் ஆங்கில ஆட்சிக்கு எதிராக மக்களைப் பொங்கி எழச் செய்தார். அந்நிய அடக்குமுறையை எதிர்த்து மேடைகளில் ஆவேசமாக உரையாற்றினார். காந்தியடிகள் சென்னையில் ஆற்றிய உரையை அற்புதமாக மொழிபெயர்த்து காந்தியடிகளிடம் பாராட்டு பெற்றார். இவருடைய அரசியல் குரு திலகர் ஆவார்.

தமது பேச்சால் தமிழ் வளர்த்தவர். திரு.வி.க. நடை என்றே ஒரு தனிநடையை நடைமுறைப்படுத்தும் அளவுக்குப் பேசுவது போலவே எழுதுவது; எழுதுவது போலவே பேசுவது என்னும் முயற்சியில் வெற்றி கண்டவர். பிணிக்கும் தகைவாய்க் கேளாரும் விரும்பும் வண்ணம் பேசுவதில் வல்லவர். அறிஞர் அண்ணா உள்ளிட்ட அக்கால இளைஞர்களை உணர்ச்சிமிகு பேச்சினால் தம்பால் ஈர்த்து மேடைத்தமிழின் முன்னோடியாகத் திகழ்ந்த வர். திரு.வி.க.

முடிவுரை :

செய்யுள் நூல்கள், உரைநடை நூல்கள் எனப் பல நூல்களை இயற்றியுள்ளார். தேசபக்தன் , நவசக்தி என்னும் இதழ்களின் வாயிலாகத் தொழிலாளர் முன்னேற்றம் பெறப் பாடுபட்டவர்.

மொழியோடு விளையாடு

இடைச்சொல் ‘கு’ சேர்த்துத் தொடரை எழுதுக.

(எ.கா.) வீடு சென்றான் – வீடு + கு + சென்றான் – வீட்டுக்குச் சென்றான்

Question 1.
மாடு புல் கொடுத்தார்
Answer:
மாடு + கு + புல் கொடுத்தார் – மாட்டுக்குப் புல் கொடுத்தார்.

Question 2.
பாட்டு பொருள் எழுது
Answer:
பாட்டு + கு + பொருள் எழுது – பாட்டுக்குப் பொருள் எழுது.

Question 3.
செடி பாய்ந்த நீர்
Answer:
செடி + கு + பாய்ந்த நீர் – செடிக்குப் பாய்ந்த நீர்.

Question 4.
முல்லை தேர் தந்தான் பாரி.
Answer:
முல்லை + கு + தேர் தந்தான் பாரி – முல்லைக்கு தேர் தந்தான் பாரி.

Question 5.
சுவர் சாந்து பூசினான்.
Answer:
சுவர் + கு + சாந்து பூசினான். – சுவருக்குச் சாந்து பூசினான்.

இரண்டு சொற்களை இணைத்துப் புதிய சொற்களை உருவாக்குக

Answer:

அகம் என முடியும் சொற்களை எழுதுக

Answer:

(எ.கா.) நூலகம்

1. உணவகம்
2. காப்பகம்
3. மருந்தகம்
4. இனிப்பகம்
5. அடுக்ககம்

கோடிட்ட இடங்களைத் தமிழ் எண் கொண்டு நிரப்புக
(எ.கா) திருக்குறள் ங பால்களைக் கொண்டது

Answer:
1. எனது வயது கஉ
2. நான் படிக்கும் வகுப்பு எ
3. தமிழ் இலக்கணம் ரு வகைப்படும்.
4. திருக்குறளில் கந அதிகாரங்கள் உள்ளன.
5. இந்தியா க கூ ச எ ஆம் ஆண்டு விடுதலை பெற்றது.

குறிப்புகளைக் கொண்டு தலைவர்களின் பெயர்களைக் கட்டங்களிலிருந்து கண்டுபிடித்து எழுதுக.

Answer:

1. மூதறிஞர் – இராஜாஜி
2. வீரமங்கை – வேலுநாச்சியார்
3. பாஞ்சாலங்குறிச்சி வீரன் – கட்டபொம்மன்
4. வெள்ளையரை எதிர்த்த தீரன் – சின்னமலை
5. கொடி காத்தவர் – திருப்பூர் குமரன்
6. எளிமையின் இலக்கணம் – காமராசர்
7. தில்லையாடியின் பெருமை – வள்ளியம்மை
8. கப்பலோட்டிய தமிழர் – சிதம்பரனார்
9. பாட்டுக்கொரு புலவன்- பாரதியார்
10. விருதுப்பட்டி வீரர் – காமராசர்
11. கள்ளுக்கடை மறியல் பெண்மணி – நாகம்மை
12. மணியாட்சியின் தியாகி – வாஞ்சிநாதன்

நிற்க அதற்குத் தக

என் பொறுப்புகள்

1. விடுதலைப் போராட்ட வீரர்களின் வரலாறுகளை அறிந்து போற்றுவேன்.
2. தலைவர்களின் அரிய பண்புகளை உணர்ந்து பின்பற்றுவேன்.

கலைச்சொல் அறிவோம்

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.3

10th Maths Exercise 1.3 Samacheer Kalvi Question 1.
Let f = {(x, y)|x, y ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Solution:
F = {(x, y)|x, y ∈ N and y = 2x}
x = {1, 2, 3,…}
y = {1 × 2, 2 × 2, 3 × 2, 4 × 2, 5 × 2 …}
R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10),…}
Domain of R = {1, 2, 3, 4,…},
Co-domain = {1, 2, 3…..}
Range of R = {2, 4, 6, 8, 10,…}
Yes, this relation is a function.

Exercise 1.3 Class 10 Maths Samacheer Question 2.
Let X = {3, 4, 6, 8}. Determine whether the relation R = {(x, f(x))|x ∈ X, f(x) = x² + 1} is a function from X to N ?
Solution:
x = {3,4, 6, 8}
R = ((x, f(x))|x ∈ X, f(x) = X² + 1}
f(x) = x² + 1
f(3) = 3² + 1 = 10
f(4) = 4² + 1 = 17
f(6) = 6² + 1 = 37
f(8) = 8² + 1 = 65

R = {(3, 10), (4, 17), (6, 37), (8, 65)}
R = {(3, 10), (4, 17), (6, 37), (8, 65)}
Yes, R is a function from X to N.

Ex 1.3 Class 10 Samacheer Question 3.
Given the function
f : x → x² – 5x + 6, evaluate
(i) f(-1)
(ii) f(2 a)
(iii) f(2)
(iv) f(x – 1)
Answer:
f(x) = x² – 5x + 6
(i) f (-1) = (-1)² – 5 (-1) + 6 = 1 + 5 + 6 = 12
(ii) f (2a) = (2a)² – 5 (2a) + 6 = 4a² – 10a + 6
(iii) f(2) = 2² – 5(2) + 6 = 4 – 10 + 6 = 0
(iv) f(x – 1) = (x – 1)² – 5 (x – 1) + 6
= x² – 2x + 1 – 5x + 5 + 6
= x² – 7x + 12

10th Maths Relation And Function Question 4.
A graph representing the function f(x) is given in figure it is clear that f(9) = 2.

(i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
(ii) For what value of x is f (x) = 1?
(iii) Describe the following
(i) Domain
(ii) Range.
(iv) What is the image of 6 under f?
Solution:
From the graph
(a) f(0) = 9
(b) f(7) = 6
(c) f(2) = 6
(d) f(10) = 0
(ii) At x = 9.5, f(x) = 1
(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {x |0 < x < 10, x ∈ R}
Range = {x|0 < x < 9, x ∈ R}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(iv) The image of 6 under f is 5.

10th Maths Exercise 1.3 Question 5.
Let f(x) = 2x + 5. If x ≠ 0 then find \(\frac{f(x+2)-f(2)}{x}\)
Solution:
Given f(x) = 2x + 5, x ≠ 0.

Samacheer Kalvi 10th Maths Exercise 1.3 Question 6.
A function fis defined by f(x) = 2x – 3
(i) find \(\frac{f(0)+f(1)}{2}\)
(ii) find x such that f(x) = 0.
(iii) find x such that f(x) = x.
(iv) find x such that f(x) = f(1 – x).
Solution:
Given f(x) = 2x – 3
(i) find \(\frac{f(0)+f(1)}{2}\)
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = -1
∴ \(\frac{f(0)+f(1)}{2}=\frac{-3-1}{2}=\frac{-4}{2}\) = -2

(ii) f(x) = 0
⇒ 2x – 3 = 0
2x = 3
x = \(\frac{3}{2}\)

(iii) f(x) = x
⇒ 2x – 3 = x ⇒ 2x – x = 3
x = 3

(iv) f(x) = f(1 – x)
2x – 3 = 2(1 – x) – 3
2x – 3 = 2x – 2x – 3
2x + 2x = 2 – 3 + 3
4x = 2
x = \(\frac{2}{4}\)
x = \(\frac{1}{2}\)

10 Std Maths Exercise 1.3 VQuestion 7.
An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown in figure. Express the volume V of the box as a function of x.

Solution:
Volume of the box = Volume of the cuboid
= l × b × h cu. units
Here l = 24 – 2x
b = 24 – 2x
h = x
∴ V = (24 – 2x) (24 – 2x) × x
= (576 – 48x – 48x + 4x²)x
V = 4x³ – 96x² + 576x

10 Maths Exercise 1.3 Question 8.
A function f is defined bv f(x) = 3 – 2x . Find x such that f(x2) = (f(x))2.
Solution:
f(x) = 3 – 2x
f(x²) = 3 – 2x²

10th Maths Exercise 1.3 In Tamil Question 9.
A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time r in hours.
Answer:
Speed of the plane = 500 km/hr
Distance travelled in “t” hours
= 500 × t (distance = speed × time)
= 500 t

10th Maths 1.3 Question 10.
The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length(x) as y = ax + b, where a, b are constants.

(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.
Solution:
(i) Given y = ax + b …………. (1)
The ordered pairs are R = {(35, 56) (45, 65) (50, 69.5) (55, 74)}
∴ Hence this relation is a function.

Substituting a = 0.9 in (2) we get
⇒ 65 = 45(.9) + b
⇒ 65 = 40.5 + b
⇒ b = 65 – 40.5
⇒ b = 24.5
∴ a = 0.9, b = 24.5
∴ y = 0.9x + 24.5
(iii) Given x = 40 , y = ?
∴ (4) → y = 0.9 (40) + 24.5
⇒ y = 36 + 24.5
⇒ y = 60.5 inches
(iv) Given y = 53.3 inches, x = ?
(4) → 53.3 = 0.9x + 24.5
⇒ 53.3 – 24.5 = 0.9x
⇒ 28.8 = 0.9x
⇒ x = \(\frac{28.8}{0.9}\) = 32 cm
∴ When y = 53.3 inches, x = 32 cm

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

12th Maths Exercise 1.1 Answers Question 1.
Find the adjoint of the following:

Solution:

Exercise 1.1 Class 12 Maths State Board Question 2.
Find the inverse (if it exists) of the following:

Solution:
For a matrix A, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}(\mathrm{adj} \mathrm{A})\). Where |A| ≠ 0. If |A| = 0 then A is called a singular matrix and so \(\mathrm{A}^{-1}\) does not exist.

12th Maths Exercise 1.1 Question 3.
If F(α) = \(\left[\begin{array}{ccc}{\cos \alpha} & {0} & {\sin \alpha} \\ {0} & {1} & {0} \\ {-\sin \alpha} & {0} & {\cos \alpha}\end{array}\right]\) show that \([\mathrm{F}(\alpha)]^{-1}=\mathrm{F}(-\alpha)\)
Solution:
Let A = F (α)
So \([\mathrm{F}(\alpha)]^{-1}=\mathrm{A}^{-1}\)
Now

12th Maths Chapter 1 Exercise 1.1 Question 4.
If A = \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]\) show that A² – 3A – 7I₂ = O₂. Hence find A^-1.
Solution:
A = \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]\)

To Find A-1
Now we have proved that A² – 3A – 7I₂ = O₂
Post multiply by A^-1 we get
A – 3I – 7A^-1 = O₂

12th Maths 1.1 Exercise Question 5.
If \(\mathbf{A}=\frac{1}{9}\left[\begin{array}{ccc}{-8} & {1} & {4} \\ {4} & {4} & {7} \\ {1} & {-8} & {4}\end{array}\right]\) prove that A^-1 = A^T
Solution:

12 Maths Exercise 1.1 Question 6.
If \(\mathbf{A}=\left[\begin{array}{rr}{8} & {-4} \\ {-5} & {3}\end{array}\right]\), verify that A(adj A) = (adj A)A = |A| I₂
Solution:

12th Maths 1st Chapter Exercise 1.1 Question 7.
If \(\mathbf{A}=\left[\begin{array}{ll}{3} & {2} \\ {7} & {5}\end{array}\right]\), and \(\mathbf{B}=\left[\begin{array}{cc}{-1} & {-3} \\ {5} & {2}\end{array}\right]\) verify that (AB)^-1 = B^-1 A^-1.
Solution:

12th Maths Application Of Matrices And Determinants Question 8.
If adj (A) = \(\left[\begin{array}{ccc}{2} & {-4} & {2} \\ {-3} & {12} & {-7} \\ {-2} & {0} & {2}\end{array}\right]\) find A
Solution:

12th Maths Ex 1.1 Question 9.
If adj(A) = \(\left[\begin{array}{ccc}{0} & {-2} & {0} \\ {6} & {2} & {-6} \\ {-3} & {0} & {6}\end{array}\right]\) find A^-1
Solution:

12th Exercise 1.1 Question 10.
Find adj(adj(A)) if adj A = \(\left[\begin{array}{ccc}{1} & {0} & {1} \\ {0} & {2} & {0} \\ {-1} & {0} & {1}\end{array}\right]\)
Solution:

12th Maths Exercise 1.1 Answers In Tamil Medium Question 11.

Solution:

12th Maths 1st Chapter Question 12.
Find the matrix A for which A \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]\)
Solution:
Given A \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]\)
Let \(\mathrm{B}=\left(\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right) \text { and } \mathrm{C}=\left(\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right)\)
Given AB = C, To find A
Now AB = C
Post multiply by B^-1 on both sides
ABB^-1 = CB^-1 (i.e) A (BB^-1) = CB^-1
⇒ A(I) = CB^-1 (i.e) A = CB^-1
To find B^-1:

12th Maths 1.1 Question 13.
Given \(\mathbf{A}=\left[\begin{array}{cc}{1} & {-1} \\ {2} & {0}\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}{3} & {-2} \\ {1} & {1}\end{array}\right] \text { and } \mathbf{C}\left[\begin{array}{ll}{1} & {1} \\ {2} & {2}\end{array}\right]\), find a matrix X such that AXB = C.
Solution:
A × B = C
Pre multiply by A^-1 and post multiply by B^-1 we get
A^-1 A × BB^-1 = A^-1CB^-1 (i.e) X = A^-1CB^-1

12 Maths Samacheer Kalvi Solutions Question 14.

Solution:

12th Maths Exercise 1.1 5th Sum Question 15.
Decrypt the received encoded message \(\left[\begin{array}{cc}{2} & {-3}\end{array}\right]\left[\begin{array}{ll}{20} & {4}\end{array}\right]\) with the encryption matrix \(\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]\) and the decryption matrix as its inverse, where the system of codes are described by the numbers 1-26 to the letters A- Z respectively, and the number 0 to a blank space.
Solution:
Let the encoding matrix be \(\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]\)


So the sequence of decoded matrices is [8 5], [12 16].
Thus the receivers read this message as HELP.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1 Additional Problems

12th Maths Chapter 1 Question 1.
Using elementary transformations find the inverse of the following matrix
Solution:

12th Maths Guide Question 2.
Using elementary transformations find the inverse of the matrix
Solution:

12th Maths Solutions Samacheer Kalvi Question 3.
Using elementary transformation find the inverse of the matrix
Solution:

12 Maths Chapter 1 Exercise 1.1 Question 4.
Using elementary transformations find the inverse of the matrix
Solution:

12th Maths Exercise 1.1 Solutions Question 5.
Using elementary transformation, find the inverse of the following matrix
Solution:

Samacheer Kalvi 12th Maths Guide Question 6.

Solution:

Samacheer Kalvi 12 Maths Solutions Question 7.

Solution:

12 Maths Solutions Samacheer Kalvi Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

You can Download Samacheer Kalvi 9th Social Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Social Science Economics Solutions Chapter 2 Employment in India and Tamilnadu

Employment in India and Tamilnadu Textual Exercise

I. Choose the correct answer.

Employment In India And Tamilnadu Question 1.
We take age group …….. years for computation of the workforce.
(a) 12 – 60
(b) 15 – 60
(c) 21 – 65
(d) 5 – 14
Answer:
(b) 15 – 60

Employment Pattern Changes Due To Answer Question 2.
Which is the correct sequence of various sectors in GDP of India in the descending order?
(a) Primary sector, Secondary sector, Tertiary sector
(b) Primary sector, Tertiary sector, Secondary sector
(c) Tertiary sector, Secondary sector, Primary sector
(d) Secondary sector, Tertiary sector, Primary sector
Answer:
(a) Primary sector, Secondary sector, Tertiary sector

Employment Pattern Changes Due To Question 3.
Which one of the following sectors is the largest employer in India?
(a) Primary Sector
(b) Secondary Sector
(c) Tertiary Sector
(d) Public sector
Answer:
(a) Primary Sector

Employment Pattern Changes Due To Dash Question 4.
Which one of the following is not in Primary Sector?
(a) Agriculture
(b) Manufacturing
(c) Mining
(d) Fishery
Answer:
(b) Manufacturing

Question 5.
Which one of the following is not in the Secondary Sector?
(a) Construction
(b) Manufacturing
(c) Small Scale Industry
(d) Forestry
Answer:
(d) Forestry

Question 6.
Tertiary Sector include/s …..
(a) Transport
(b) Insurance
(c) Banking
(d) All of these
Answer:
(d) All of these

Question 7.
Match the List I with List II using the codes given below:

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

Question 8.
Which sector is not included in the occupational pattern?
(a) Primary sector
(b) Secondary sector
(c) Tertiary sector
(d) Private sector
Answer:
(d) Private sector

Question 9.
Which Delhi Sultan of medieval India formed ‘Employment Bureau’ to solve the un-employment problem?
(a) Muhamad Bin Tugluq
(b) Allauddin Khilji
(c) Feroz Shah Tugluq
(d) Balban
Answer:
(c) Feroz Shah Tugluq

Question 10.
…….. sector is registered and follows government rules.
(a) Agriculture
(b) Organised
(c) Unorganised
(d) Private
Answer:
(b) Organised

Question 11.
……. sector provides j.ob security and higher wages
(a) Public sector
(b) Organised sector
(c) Unorganised sector
(d) Private sector
Answer:
(b) Organised sector

Question 12.
Find the odd one.
(a) Banking
(b) Railways
(c) Insurance
(d) Small Scale Industry
Answer:
(b) Railways

Question 13.
The sectors are classified into Public and Private sectors on the basis of …
(a) number of workers employed
(b) nature of economic activity
(c) ownership of enterprises
(d) employment conditions
Answer:
(c) ownership of enterprises

Question 14.
Assertion (A): The unorganised sector of the economy characterised by the household manufacturing activity and small-scale industry.
Reason (R): Jobs here are low paid and often not regular
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A)
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 15.
People who employ workers and pay rewards for their work is termed as ……
(a) employee
(b) employer
(c) labour
(d) caretaker
Answer:
(b) employer

Question 16.
……. continues to be the largest employer in Tamil Nadu.
(a) Agriculture
(b) Manufacturing
(c) Banking
(d) Small Scale Industry
Answer:
(a) Agriculture

II. Fill in the blanks.

1. In ……. sector, the employment terms are not fixed and regular.
2. Economic activities are classified into ……….. and …… sectors.
3. ……… has always featured as an important element of development policy in India.
4. Employment pattern changes due to …….
5. The nature of employment in India is ………
6. ……. of the economy is the number of people in the country, who work and also
capable of working.
7. Public sector means ……… .
Answers:
1. unorganised
2. Public and Private
3. Employment
4. life style of the people
5. multi-dimensional
6. Labour force
7. Government undertaking

III. Match the following.

Answers:
1. (d)
2. (c)
3. (b)
4. (a)

IV. Give short answers.

Question 1.
What is labour force of the economy?
Answer:
Labour force of the economy is the number of people in the country who work and also capable of working.

Question 2.
Why are children and old age (above 60 years) are not considered for computation of workforce?
Answer:
We take the age group of 15-60 years for the computation of work force. Persons who are less than 15 years are considered as children and person who have crossed 60 years of age are excluded as they are not physically fit to undertake productive occupation. Hence they are not considered for computation of work force.

Question 3.
What are the three sectors of an economy?
Answer:
The economy is classified into three sectors. Primary (or) Agriculture sector, Secondary (or) Industrial sector and Tertiary (or) Service sector.

Question 4.
Agriculture, despite a sharp decline in Gross Domestic Product, continues to be the largest employer in Tamil Nadu. Give reason.
Answer:
Agriculture, despite a sharp decline in gross domestic product, continues to be the largest employer in Tamil Nadu. This is because the non-agriculture sectors are yet to generate enough employment to affect a shift of labour force.

V. Answer in detail.

Question 1.
Explain:
(a) primary sector;
(b) secondary sector;
(c) tertiary sector.
Answer:
The structure of employment denotes the number of workers engaged in different sectors of the economy.
Primary Sector: Agriculture, forestry, animal husbandry, poultry, dairy farming, fishing, etc. Secondary Section: Manufacturing, small and large scale industries and constructional activities. .
Tertiary Sector: Transport insurance banking, trade, communication, real estate, government and non-government services.
In developing countries like India that a large work force will be engaged in Primary Sector, while a small proportion in Secondary and Tertiary sectors.

Question 2.
Explain the employment structure of India.
Answer:

1. The nature of employment in India is multi-dimensional. Some get employment throughout the year; some others get employed for only a few months in a year.
2. The economy is classified into three sectors: primary or agriculture sector, secondary or industrial sector and tertiary or service sector.
3. The structure of employment denotes the number of workers engaged in different sectors of the economy.
4. Though the occupational pattern varies from one country to another, one can find in developing countries like India that a large work force will be engaged in primary sector, while a small proportion in secondary and tertiary sectors.
5. Whereas, in well-developed countries, the proportion of workforce engaged in agriculture will be very small and a majority of labour force will be in the industrial and tertiary sectors.
6. Employment has always featured as an important element of development policy in India.
7. Employment growth has increased at an average rate of 2% during the past four decades since 1972 – 73.

Question 3.
Compare the employment conditions prevailing in the organised and unorganised
sectors.
Answer:

Question 4.
Distinguish between the Public sector and the Private sector.
Answer:

VI. Projects and Activities.

Question 1.
Make a long list of all kinds of work that you find for adults around you. In what way can you classify them?
Answer:
(a) Agriculture, Food & Natural resources.
(b) Business, Management & Administration.
(c) Communication & Information Systems.
(d) Engineering, Manufacturing & Technology.
(e) Health Science Technology.
(f) Human Services.

Question 2.
A research scholar looked at the working people in the city of Chennai and found the following:
Answer:

Question 3.
Classify the following list of occupations under primary, secondary and tertiary sectors. Milk vendor, tailor, teacher, doctor, farmer, postman, engineer, potter, fisherman, artisans, policeman, banker, driver, carpenter.
Answer:

VII. HOTS.

Tertiary sector is in top position in the world now. Justify
Answer:

1. In a direct sense, the tertiary sector is mostly independent.
2. This sector primarily includes those activities that help in the development of the primary and secondary sector.
3. They aid and support the production process.
   e.g., The transport sector, that uses trucks to transport finished products from factory to the retail stores.
4. Service sector also includes some essential services that may not directly help in the production of goods.
   e.g., Teachers, doctors, washermen, barbers, etc.
5. Tertiary sector actually caters to the excess income in a person’s hand after he is successful in acquiring his basic needs.

This sector will remain a chief engine for future growth.

VIII. Life Skill.

Question 1.
Discuss the sectors of your village economy.
Answer:

1. The teacher will make the students to gain knowledge of Primary, Secondary and Tertiary sector.
2. The students will be asked to collect information regarding the three sectors.
3. The students will be advised to classify the jobs available under Primary, Secondary and Tertiary sectors in their village.

Employment in India and Tamilnadu Additional Questions

I. Choose the correct answer.

Question 1.
The nature of employment in India is
(a) Two-dimensional
(b) Three dimensional
(c) Multi-dimensional
(d) All the above
Answer:
(c) Multi-dimensional

Question 2.
This is Primary Sector.
(a) Agriculture
(b) Manufacturing
(c) Small scale Industries
(d) Banking
Answer:
(a) Agriculture

Question 3.
This sector do not enjoy any special benefit.
(a) Organised Sector
(b) Unorganised Sector
(c) Public Sector
(d) Private Sector
Answer:
(b) Unorganised Sector

Question 4.
Assertion (A): Unorganised sector jobs are low paid.
Reason (R): These employees have no job-security.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A).
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(b) Both (A) and (R) are true and (R) does not explain (A).

Question 5.
Find the odd one.
(a) Transport
(b) Insurance
(c) Banking
(d) Poultry
Answer:
(d) Poultry

Question 6.
This is Private Sector.
(a) TVS Motors
(b) NLC
(c) SAIL
(d) BSNL
Answer:
(a) TVS Motors

II. Fill in the blanks.

1. To survive in the world we all need …… to earn money.
2. In well developed countries the proportion of work force engaged in ……. will be very smart.
3. …….. sector has some formal processes and procedures.
4. …….. are not registered with the government.
5. Most of the …….. in Tamil Nadu has been contributed by the unorganised and informal sectors.
Answers:
1. employment
2. Agriculture
3. Organized
4. Unorganised Sectors
5. employment growth

III. Match the following.

Answers:
1. (d)
2. (a)
3. (c)
4. (b)

IV. Give short answers.

Question 1.
Who are called employees ?
Answer:
Those who are engaged in economic activities, in whatever capacity – high (or) low are called employees.

Question 2.
What is the nature of employment in India?
Answer:
The nature of employment in India is multi-dimensional. Some get employment throughout the year. Some others get employed for only a few months in a year.

Question 3.
Mention the recent trends in the working pattern of the employees.
Answer:
The trends are
(a) Increasing self-employment.
(b) Firms using fewer full-time employees and tending to offer more short-term contracts.
(c) There has been a growth in part-time employment.

Question 4.
What does structure of employment denote?
Answer:
Structure of employment denotes number of workers engaged in different sectors of the economy.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.5

10th Maths Exercise 3.5 Samacheer Kalvi Question 1.
Simplify

Solution:

Ex 3.5 Class 10 Samacheer Question 2.
Simplify

Solution:

10th Maths Exercise 3.5 Samacheer Question 3.
Simplify

Solution:

10th Maths Exercise 3.5 Question 4.
If x = \(\frac{a^{2}+3 a-4}{3 a^{2}-3}\) and y = \(\frac{a^{2}+2 a-8}{2 a^{2}-2 a-4}\) find the value of x²y².
Solution:

Exercise 3.5 Class 10 Maths Samacheer Question 5.
If a polynomial p(x) = x² – 5x – 14 is divided by another polynomial q(x) we get \(\frac{x-7}{x+2}\) find q(x).
Solution:
p(x) = x² – 5x – 14

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Resolve the following rational expressions into partial fractions

11th Maths Exercise 2.9 Answers Question 1.

Solution:

11th Maths Exercise 2.9 Question 2.

Solution:

Exercise 2.9 Class 11 Question 3.

Solution:

11th Std Maths Exercise 2.9 Solutions Question 4.

Solution:

11th Maths Exercise 2.9 Solutions Question 5.

Solution:

Equating nuemarator on bothsides we get

Ex 2.9 Class 11 Maths Question 6.

Solution:

Equating numerator on both sides
(x – 2)² = A(x² + 1) + (Bx + c)(x)
Put x = 0
1 = A
Equating co-eff of x²
1 = A + B
(i.e.,) 1 + B = 1 ⇒ B = 0
put x = 1
A(2) + B + C = 0 (i.e.,) 2A + B + C = 0
2 + 0 + C = 0 ⇒ C = -2

11 Maths Exercise 2.9 Question 7.

Solution:
Since numerator and denominator are of same degree
we have divide the numerator by the denominator

Substituting the value in ….(1)

Question 8.

Solution:
Numerator is of greater degree than the denominator
So dividing Numerator by the denominator

⇒ 21x + 31 = A(x + 3) + B(x + 2)
Put x = -3
-63 + 31 = B(-1)
B = 32
Put x = -2
-42 + 31 = A(1) + B(0)
A = -11

11th Maths Exercise 2.9 3rd Sum Question 9.

Solution:

11th Maths Exercise 2.9 5th Sum Question 10.

Solution:

Equating Numerator on both sides we get
6x² – x + 1 = A(x² + 1) + (Bx + c)(x + 1)
6 + 1 + 1 = A(2) + 0 ⇒ 2A = 8 ⇒ A = 4
Equating co-eff of x²
6 = A + B
(i.e.,) 4 + B = 6 ⇒ B = 6 – 4 = 2
put x = 0
1 = A+ C
4 + C = 1 ⇒ C = 1 – 4 = -3

Samacheer Kalvi 11th Maths Example Sums Question 11.

Solution:
Since Numerator and are of same degree divide Numerator by the denominator

equating Numerator on both sides we get
x – 5 = A(x + 3) + B(x – 1)
Put x = -3
-3 -5 = A(0) + B(-4)
-4B = -8 ⇒ B = 2
Put x = 1
1 – 5 = A(4) + B(0)
4A = -4 ⇒ A = -1

Samacheer Kalvi Guru 11th Maths Question 12.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 Additional Questions

Samacheer Kalvi 11 Maths Solutions Question 1.

Solution:

Samacheer Kalvi 11th Maths Question 2.

Solution:

Samacheer Kalvi 11th Maths Solution Question 3.

Solution:

Equating nemerator on b/s
9 = A(x+2)² + B(x – 1)(x + 2) + C(x – 1)
Put x = -2
9 = A(0) + B(0) + C(-3)
-3C = 9 ⇒ C = -3
Put x = 1
9 = A (1 + 2)² + B (0) + C(0)
9A = 9
A = 1
Put x = 0
9 = 4A – 2B – C
9 = 4(1) – 2B + 3
9 – 7 = -2B
2 = -2B
B = -1

Samacheer Kalvi 11th Maths Book Solutions Question 4.

Solution:

11 Maths Solutions Samacheer Question 5.

Solution:

0 = 0 + B(1 + 2)
3B = 0 ⇒ B = 0
Put x = -2
(-2)³ – 1 = A(-2 – 1) + B(0)
-8 – 1 = -3A
-9 = -3A
A = 9/3 ⇒ A = 3

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.3

10th Maths Exercise 3.3 Samacheer Kalvi Question 1.
Find the LCM and GCD for the following and verify that f(x) × g(x) = LCM × GCD.
(i) 21x²y, 35xy²
(ii) (x³ – 1)(x + 1), x³ + 1
(ii) (x³ – 1) (x + 1), (x³ – 1)
(iii) (x²y + xy²), (x² + xy)
Solution:
(i) f(x) = 21x²y = 3 × 7x²y
g(x) = 35xy² = 7 × 5xy²
G.C.D. = 7xy
L.C.M. = 7 × 3 × 5 × x²y² = 105x² × y²
L.C.M × G.C.D = f(x) × g(x)
105x²y² × 7xy = 21x²y × 35xy²
735x³y³ = 735x³y³
Hence verified.

(ii) (x³ – 1)(x + 1) = (x – 1)(x² + x + 1)(x + 1)
x³ + 1 = (x + 1) (x² – x + 1)
G.C.D = (x+ 1)
L.C.M = (x – 1)(x + 1)(x² + x + 1)(x² – x + 1)
∴ L.C.M. × G.C.D = f(x) × g(x)
(x – 1)(x + 1)(x² + x + 1) (x² – x + 1) = (x – 1)
(x² + x + 1) × (x + 1) (x² – x + 1)
(x³ – 1)(x + 1)(x³ + 1) = (x³ – 1)(x + 1)(x³ + 1)
∴ Hence verified.

(iii) f(x) = x²y + xy² = xy(x + y)
g(x) = x² + xy = x(x + y)
L.C.M. = x y (x + y)
G.C.D. = x (x + y)
To verify:
L.C.M. × G.C.D. = xy(x + y) × (x + y)
= x²y (x + y)² ……….. (1)
f(x) × g (x) = (x²y + xy²)(x² + xy)
= x²y (x + y)² …………… (2)
∴ L.C.M. × G.C.D = f(x) × g{x).
Hence verified.

Ex 3.3 Class 10 Samacheer Question 2.
Find the LCM of each pair of the following polynomials
(i) a² + 4a – 12, a² – 5a + 6 whose GCD is a – 2
(ii) x⁴ – 27a³x, (x – 3a)² whose GCD is (x – 3a)
Solution:
(i) f(x) = a² + 4a – 12 = (a + 6)(a – 2)

(ii) f(x) = x⁴ – 27a³x = x(x³ – (3a)³)
g(x) = (x – 3a)²
G.C.D = (x – 3a)
L.C.M. × G.C.D = f(x) × g(x)
L. C.M = \(\frac{x\left(x^{3}-(3 a)^{3}\right) \times(x-3 a)^{2}}{(x-3 a)}\)
L.C.M = x(x³ – (3a)³) . (x – 3a)
= x(x – 3a)² (x² + 3ax + 9a²)

10th Maths Exercise 3.3 Question 3.
Find the GCD of each pair of the following polynomials
(i) 12(x⁴ – x³), 8(x⁴ – 3x³ + 2x²) whose LCM is 24x³ (x – 1)(x – 2)
(ii) (x³ + y³), (x⁴ + x²y² + y⁴) whose LCM is (x³ + y³) (x² + xy + y²)
Solution:
(i) f(x)= 12(x⁴ – x³)
g(x) = 8(x⁴ – 3x³ + 2x²)
L.C.M = 24x³ (x – 1)(x – 2)

(ii) (x³ + y³), (x⁴ + x²y² + y⁴)
L.C.M. = (x³ + y³)(x² + xy + y²)

Exercise 3.3 Class 10 Maths Samacheer Question 4.
Given the LCM and GCD of the two polynomials p(x) and q(x) find the unknown polynomial in the following table

Solution: