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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.2

10th Maths Exercise 3.2 Samacheer Kalvi Question 1.
Find the GCD of the given polynomials
(i) x⁴ + 3x³ – x – 3, x³ + x² – 5x + 3
(ii) x⁴ – 1, x³ – 11x² + x – 11
(iii) 3x⁴ + 6x³ – 12x⁴ – 24x, 4x⁴ + 14x³ + 8x² – 8x
(iv) 3x³ + 3x² + 3x + 3, 6x³ + 12x² + 6x + 12
Solution:
x⁴ + 3x³ – x – 3, x³ + x² – 5x + 3
Let f(x) = x⁴ + 3x³ – x – 3
g(x) = x³ + x² – 5x + 3

Note that 3 is not a divisor of g(x). Now dividing g(x) = x³ + x² – 5x + 3 by the new remainder x² + 2x – 3 (leaving the constant factor 3) we get

Here we get zero remainder
G.C.D of (x⁴ + 3x³ – x – 3), (x³ + x² – 5x + 3) is (x² + 2x – 3)

(ii) x⁴ – 1, x³ – 11x² + x – 11

(iii) 3x⁴ + 6x³ – 12x² – 24x, 4x⁴ + 14x³ + 8x² – 8x
4x⁴ + 14x³ + 8x² – 8x = 2 (2x⁴ + 7x³ + 4x² -4x)
Let us divide
(2x⁴ + 7x³ + 4x² + 4x) by x⁴ + 2x³ – 4x² – 8x

(x³ + 4x³ + 4x) ≠ 0
Now let us divide
x⁴ + 2x³ – 4x² – 8x by x³ + 4x² + 4x

∴ x³ + 4x² + 4x is the G.C.D of 3x⁴ + 6x³ -12x² – 24x, 4x⁴ + 14x³ + 8x² -8x
∴ Ans x (x² + 4x + 4)

(iv) f(x) = 3x³ + 3x² + 3x + 3 = 3(x³ + x² + x + 1)
g(x) = 6x³ + 12x² + 6x + 12
= 6(x³ + 2x² + x + 2)
= 2 × 3 (x³ + 2x² + x + 2)
f(x) ⇒ x³ + x² + x + 1

Ex 3.2 Class 10 Samacheer Question 2.
Find the LCM of the given expressions,
(i) 4x²y, 8x³y²
(ii) -9a³b², 12a²b²c
(iii) 16m, -12m²n², 8n²
(iv) p² – 3p + 2, p² – 4
(v) 2x² – 5x – 3, 4x² – 36
(vi) (2x² – 3xy)², (4x – 6y)³, 8x³ – 27y³
Solution:
(i) 4x²y, 8x³y²
4x²y = 2 × 2 x²y
8x³y² = 2 × 2 × 2 x³y²
L.C.M. = 2 × 2 × 2 x³y²
= 8x³ y²

(ii) -9a³b² = -3 × 3 a³b²
12a²b²c = 2 × 3 × 2a²b²c
L.C.M. = -3 × 3 × 2 × 2 a³b²c
= -36a³b²c

(iii) 16m, -12m²n², 8n²
16 m = 2 × 2 × 2 × 2 × m
-12m²n² = -2 × 2 × 3 × m²n²
8n² = 2 × 2 × 2 × n²
L.C.M.= -2 × 2 × 2 × 2 × 3 m²n²
= -48 m²n²

(v) 2x² – 5x – 3, 4x² – 36
2x² – 5x – 3 = (x – 3)(2x + 1)
4x² – 36 = 4(x + 3)(x – 3)
L.C.M. = 4(x + 3)(x – 3)(2x + 1)

(vi) (2x² – 3xy)² = (x(2x – 3y))²
(4x – 6y)³ = (2(2x – 3y))³
8x³ – 27y³= (2x)³ – (3y)³
= (2x – 3y) (4x² + 6xy + 9y²)
L.C.M. = 2³ × x² (2x – 3y)³ (4x² + 6xy + 9y²)

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 1.4

Exercise 1.4 Class 10 Maths Samacheer Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.

Solution:

(i) It is not a function. The graph meets the vertical line at more than one points.
(ii) It is a function as the curve meets the vertical line at only one point.
(iii) It is not a function as it meets the vertical line at more than one points.
(iv) It is a function as it meets the vertical line at only one point.

10th Maths Exercise 1.4 Samacheer Kalvi Question 2.
Let f :A → B be a function defined by f(x) = \(\frac{x}{2}\) – 1, Where A = {2, 4, 6, 10, 12},
B = {0, 1, 2, 4, 5, 9}. Represent f by
(i) set of ordered pairs;
(ii) a table;
(iii) an arrow diagram;
(iv) a graph
Solution:
f: A → B
A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}

(i) Set of ordered pairs
= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
(ii) a table

(iii) an arrow diagram;

Ex 1.4 Class 10 Samacheer Question 3.
Represent the function f = {(1, 2),(2, 2),(3, 2), (4,3), (5,4)} through
(i) an arrow diagram
(ii) a table form
(iii) a graph
Solution:
f = {(1, 2), (2, 2), (3, 2), (4, 3), (5, 4)}
(i) An arrow diagram.

10th Maths Exercise 1.4 Question 4.
Show that the function f : N → N defined by f{x) = 2x – 1 is one – one but not onto.
Solution:
f: N → N
f(x) = 2x – 1
N = {1, 2, 3, 4, 5,…}
f(1) = 2(1) – 1 = 1
f(2) = 2(2) – 1 = 3
f(3) = 2(3) – 1 = 5
f(4) = 2(4) – 1 = 7
f(5) = 2(5) – 1 = 9

In the figure, for different elements in x, there are different images in f(x).
Hence f : N → N is a one-one function.
A function f: N → N is said to be onto function if the range of f is equal to the co-domain of f
Range = {1, 3, 5, 7, 9,…}
Co-domain = {1, 2, 3,..}
But here the range is not equal to co-domain. Therefore it is one-one but not onto function.

10th Maths 1.4 Exercise Question 5.
Show that the function f: N → N defined by f (m) = m² + m + 3 is one – one function.
Solution:
f: N → N
f(m) = m² + m + 3
N = {1, 2, 3, 4, 5…..}m ∈ N
f{m) = m² + m + 3
f(1) = 1² + 1 + 3 = 5
f(2) = 2² + 2 + 3 = 9
f(3) = 3² + 3 + 3 = 15
f(4) = 4² + 4 + 3 = 23

In the figure, for different elements in the (X) domain, there are different images in f(x). Hence f: N → N is a one to one but not onto function as the range of f is not equal to co-domain.
Hence it is proved.

10th Maths Ex 1.4 Question 6.
Let A = {1,2, 3, 4} and B = N .
Let f: A → B be defined by f(x) = x³ then,
(i) find the range of f
(ii) identify the type of function
Answer:
A = {1,2, 3,4}
B = {1,2, 3, 4, 5,….}
f(x) = x³
f(1) = 1³ = 1
f(2) = 2³ = 8
f(3) = 3³ = 27
f(4) = 4³ = 64
(i) Range = {1,8, 27, 64}
(ii) one -one and into function.

10th Maths Exercise 1.4 Answers Question 7.
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f: R → R defined by f(x) = 2x + 1
(ii) f: R → R defined by f(x) = 3 – 4x²
Solution:
(i) f : R → R
f(x) = 2x + 1
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(-1) = 2(-1) + 1 = -1
f(0) = 2(0) + 1 = 1
It is a bijective function. Distinct elements of A have distinct images in B and every element in B has a pre-image in A.
(ii) f: R → R; f(x) = 3 – 4x²
f(1) = 3 – 4(1²) = 3 – 4 = -1
f(2) = 3 – 4(2²) = 3 – 16 = -13
f(-1) = 3 – 4(-1)² = 3 – 4 = -1
It is not bijective function since it is not one-one

Samacheer Kalvi 10th Maths Book Graph Solution Question 8.
Let A = {-1, 1} and B = {0, 2}. If the function f: A → B defined by f(x) = ax + b is an onto function? Find a and b.
Solution:
A= {-1, 1},B = {0, 2}
f: A → B, f(x) = ax + b
f(-1) = a(-1) + b = -a + b
f(1) = a(1) + b = a + b
Since f(x) is onto, f(-1) = 0
⇒ -a + b = 0 …(1)
& f(1) = 2
⇒ a + b = 2 …(2)
-a + b = 0

10th Maths Exercise 1.4 In Tamil Question 9.
If the function f is defined by

(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
Solution:
(i) f(3) ⇒ f(x) = x + 2 ⇒ 3 + 2 = 5
(ii) f(0) ⇒ 2
(iii) f (- 1.5) = x – 1
= -1.5 – 1 = -2.5
(iv) f(2) + f(-2)
f(2) = 2 + 2 = 4     [∵ f(x) = x + 2]
f(-2) = -2 – 1 = -3    [∵ f(x) = x – 1]
f(2) + f(-2) = 4 – 3 = 1

Samacheer Kalvi 10th Maths Exercise 1.4 Question 10.
A function f: [-5,9] → R is defined as follows:

Solution:
f : [-5, 9] → R
(i) f(-3) + f(2)
f(-3) = 6x + 1 = 6(-3) + 1 = -17
f(2) = 5 × 2 – 1 = 5(2²) – 1 = 19
∴ f(-3) + f(2) = -17 + 19 = 2

(ii) f(7) – f(1)
f(7) = 3x – 4 = 3(7) – 4 = 17
f(1) = 6x + 1 = 6(1) + 1 = 7
f(7) – f(1) = 17 – 7 = 10

(iii) 2f(4) + f(8)
f(4) = 5x² – 1 = 5 × 4² – 1 = 79
f(8) = 3x – 4 = 3 × 8 – 4 = 20
∴ 2f(4) + f(8) = 2 × 79 + 20 = 178

10th Maths 1.4 Question 11
The distance S an object travels under the influence of gravity in time t seconds is 1 2 given by S(t) = \(\frac { 1 }{ 3 } \)gt² + at + b, where, (g is the acceleration due to gravity), a, b are constants. Check if the function S(t) is one-one.
Answer:
S(t) = \(\frac { 1 }{ 2 } \)gt² + at + b
Let the time be 1, 2, 3 …. n seconds
S(1) = \(\frac { 1 }{ 2 } \)g(1)² + a(1) + b
= \(\frac { g }{ 2 } \) + a + b
S(2) = \(\frac { 1 }{ 2 } \) g(2)² + a(2) + b
= \(\frac { 4g }{ 2 } \) + 2a + b
= 2g + 2a + b
S(3) = \(\frac { 1 }{ 2 } \) g(3)² + a(3) + 6
= \(\frac { 9 }{ 2 } \) g + 3a + b
For every different value of t, there will be different distance.
∴ It is a one-one function.

Samacheer Kalvi 10th Maths Book Graph Solutions Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C)= F where F = \(\frac{9}{5}\) C + 32 . Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Farenheit value.
Solution:

Students can Download Tamil Chapter 1.2 கவின்மிகு கப்பல் 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 2 Chapter 1.2 கவின்மிகு கப்பல்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
இயற்கை வங்கூழ் ஆட்ட – அடிக்கோடிட்ட சொல்லின் பொருள்
அ) நிலம்
ஆ) நீர்
இ) காற்று
ஈ) நெருப்பு
Answer:
இ) காற்று

Question 2.
மக்கள் ………………….. ஏறி வெளிநாடுகளுக்குச் சென்றனர்.
அ) கடலில்
ஆ) காற்றில்
இ) கழனியில்
ஈ) வங்கத்தில்
Answer:
ஈ) வங்கத்தில்

Question 3.
புலால் நாற்றம் உடையதாக அகநானூறு கூறுவது ……………….
அ) காற்று
ஆ) நாவாய்
இ) கடல்
ஈ) மணல்
Answer:
இ) கடல்

Question 4.
‘பெருங்கடல்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது
அ) பெரு + கடல்
ஆ) பெருமை + கடல்
இ) பெரிய + கடல்
ஈ) பெருங் + கடல்
Answer:
ஆ) பெருமை + கடல்]

Question 5.
இன்று + ஆகி என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் …….
அ) இன்று ஆகி
ஆ) இன்றி ஆகி
இ) இன்றாகி
ஈ) இன்றா ஆகி
Answer:
இ) இன்றாகி

Question 6.
எதுகை இடம்பெறாத இணை ………..
அ) இரவு – இயற்கை
ஆ) வங்கம் – சங்கம்
இ) உலகு – புலவு
ஈ) அசைவு – இசைவு
Answer:
அ) இரவு – இயற்கை

பொருத்துக

1. வங்கம் – பகல்
2. நீகான் – கப்பல்
3. எல் – கலங்கரை விளக்கம்
4. மாட ஒள்ளெரி – நாவாய் ஓட்டுபவன்
Answer:
1. வங்க ம் – கப்பல்
2. நீகான் – நாவாய் ஓட்டுபவன்
3. எல் – பகல்
4. மாட ஒள்ளெரி – கலங்கரை விளக்கம்

குறுவினா

Question 1.
நாவாயின் தோற்றம் எவ்வாறு இருந்ததாக அகநானூறு கூறுகிறது?
Answer:
நாவாயின் தோற்றம் : உலகம் புடைபெயர்ந்தது போன்ற அழகு பொருந்திய தோற்றத்தை உடையது நாவாய் என்று அகநானூறு நாவாயின் தோற்றத்தைப் பற்றி கூறுகிறது.

Question 2.
நாவாய் ஓட்டிகளுக்குக் காற்று எவ்வாறு துணை செய்கிறது?
Answer:
இரவும் பகலும் ஓரிடத்தும் தங்காமல் வீசுகின்ற காற்றானது நாவாயை அசைத்துச் செலுத்தி நாவாய் ஓட்டிகளுக்கு உறுதுணையாக நிற்கிறது.

சிறுவினா

கடலில் கப்பல் செல்லும் காட்சியை அகநானூறு எவ்வாறு விளக்குகிறது?
Answer:

• உலகம் புடைபெயர்ந்தது போன்ற அழகு பொருந்திய தோற்றத்தை உடையது நாவாய்.
•  அந்த நாவாய் புலால் நாற்றமுடைய அலைவீசும் பெரிய கடலின் நீரைப் பிளந்து கொண்டு செல்லும்.
• இரவும் பகலும் ஓரிடத்தும் தங்காமல் வீசுகின்ற கடற்காற்றானது நாவாயை அசைத்துச் செலுத்த பெரிதும் துணை புரிகின்றது.
• உயர்ந்த கரையை உடைய மணல் நிறைந்த துறைமுகத்தில் கலங்கரை விளக்கத்தின் ஒளியால் திசை அறிந்து நாவாய் ஓட்டுபவன் நாவாயைச் செலுத்துவான் என்று அகநானூறு கடலில் கப்பல் செல்லும் காட்சியை விளக்குகிறது.

சிந்தனை வினா

தரைவழிப் பயணம், கடல்வழிப் பயணம் ஆகியவற்றுள் நீங்கள் விரும்புவது எது? ஏன்?
Answer:

• நான் விரும்பும் பயணம் கடல்வழிப் பயணம்.
• ஏனென்றால் கடலில் செல்லும் போது கடலில் வீசும் இதமான காற்று உடலை வருடிச் செல்லும். மேலும் கீழுமாக தாவிச் செல்லும் அலைகள் காண்பதற்கு கவினுற அமைந்திருக்கும்.
• கடலில் வாழும் பல்வகை மீன்கள் கடல் நீரில் நீந்திச் செல்லும் காட்சி பார்ப்பதற்கு மிக அழகாக இருக்கும்.
• இரவு நேரத்தில் கப்பலிலிருந்து ஆகாயத்தைப் பார்க்கும் போது பல்வகை விண்மீன்கள் மற்றும் நிலவு நம்முடனே பயணிப்பது போன்ற தோற்றம் நம்மை மகிழ்ச்சியில் ஆழ்த்தும்.

கற்பவை கற்றபின்

Question 1.
கடலில் கிடைக்கும் பொருள்களின் பெயர்களைத் தொகுக்க.
Answer:
கடலில் கிடைக்கும் பொருட்களின் பெயர்கள் :

• பல்வகை மீன்கள், சிப்பிகள், சங்குகள், நண்டுகள் கடலின் மூலம் கிடைக்கின்றன.
• சுண்ணாம்பு. மணல், சரளை போன்ற பொருட்கள் மற்றும் கடல் அடிவாரத்தில் கரைந்துள்ள கனிமங்கள்.
• கச்சா எண்ணெய் மற்றும் எரிவாயு.
• கடல் நீரில் இருந்து உப்பு கிடைக்கிறது.
• கடல்வாழ், உயிரினங்களில் முத்துக்களை உற்பத்தி செய்யும் திறனுடைய யூனியோ, க்வாட்ருலா என்ற பெயருடைய சிப்பிகள் உள்ளன.
• ஆழ்கடலில் எடுக்கப்படும் முத்து உயர் ரகமாகும். முத்தை அணிந்தால் முத்து உடலில் பட்டு கரையும். அப்போது உடல் சூடு நீங்கும் என மருத்துவர்கள் குறிப்பிடுகின்றனர்.

Question 2.
கடற்பயணம் பற்றிய சிறுகதை ஒன்றை அறிந்து வந்து வகுப்பறையில் பகிர்க.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டியவை.

கூடுதல் வினாக்கள்

சொல்லும் பொருளும் :

1. உரு – அழகு
2. வங்கம் – கப்பல்
3. போழ – பிளக்க
4. எல் – பகல்
5. வங்கூழ் – காற்று
6. கோடு உயர் – கரை உயர்ந்த
7. நீகான் – நாவாய் ஓட்டுபவன்
8. மாட ஒள்ளெரி – கலங்கரை விளக்கம்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
‘உரு’ என்பதன் பொருள் …..
அ) நாவாய் ஓட்டுபவன்
ஆ) கப்பல்
இ) கரை உயர்ந்த
ஈ) அழகு
Answer:
:ஈ) அழகு

Question 2.
‘போழ’ என்பதன் பொருள் …………………
அ) பிளக்க
ஆ) காற்று
இ) கப்பல்
ஈ) கலங்கரை விளக்கம்
Answer:
அ) பிளக்க

Question 3.
‘வங்கூழ்’ என்பதன் பொருள் ……………….
அ) காற்று
ஆ) அழகு
இ) பிளக்க
ஈ) பகல்
Answer:
அ) காற்று

Question 4.
‘நீகான்’ என்னும் சொல்லுக்கு …………………………. என்பது பொருள்.
அ) நாவாய் ஓட்டுபவன்
ஆ) கலங்கரை விளக்கம்
இ) பகல்
ஈ) கப்பல்
Answer:
அ) நாவாய் ஓட்டுபவன்

Question 5.
‘வங்கம்’ என்பதன் பொருள் ……………….
அ) கப்பல்
ஆ) பகல்
இ) அழகு
ஈ) காற்று
Answer:
அ) கப்பல்

Question 6.
கோடு உயர் என்பதன் பொருள் ……………..
அ) கரை உயர்ந்த
ஆ) கப்பல்
இ) காற்று
ஈ) பிளக்க
Answer:
அ) கரை உயர்ந்த

Question 7.
மாட ஒள்ளெரி என்பதன் பொருள் ………………
அ) கலங்கரை விளக்கம்
ஆ) பிளக்க
இ) கரை உயர்ந்த
ஈ) காற்று
Answer:
அ) கலங்கரை விளக்கம்

விடையளி:

Question 1.
அகநானூறு எட்டுத்தொகை நூல்களுள் ஒன்றா?
Answer:
ஆம். இது எட்டுத் தொகை நூல்களுள் ஒன்று.

Question 2.
எட்டுத் தொகை நூல்களை வரிசைப்படுத்துக:
Answer:
நற்றிணை, குறுந்தொகை, ஐங்குறுநூறு, பதிற்றுப்பத்து, பரிபாடல், கலித்தொகை, அகநானூறு, புறநானூறு.

Question 3.
மருதன் இளநாகனார் குறிப்பு வரைக.
Answer:

• மருதன் இளநாகனார் சங்ககாலப் புலவர்களுள் ஒருவர்.
• கலித்தொகையின் மருதத்திணையில் உள்ள முப்பத்தைந்து பாடல்களையும் பாடியவர்.
• மருதத்திணை பாடுவதில் வல்லவர் என்பதால் மருதன் இளநாகனார் என அழைக்கப்படுகிறார்.

பாடலின் பொருள்

உலகம் புடைபெயர்ந்தது போன்ற அழகு பொருந்திய தோற்றத்தை உடையது நாவாய். அது புலால் நாற்றமுடைய அலைவீசும் பெரிய கடலின் நீரைப் பிளந்து கொண்டு செல்லும். இரவும் பகலும் ஓரிடத்தும் தங்காமல் வீசுகின்ற காற்றானது நாவாயை அசைத்துச் செல்லும். உயர்ந்த கரையை உடைய மணல் நிறைந்த துறைமுகத்தில் கலங்கரை விளக்கத்தின் ஒளியால் திசை அறிந்து நாவாய் ஓட்டுபவன் நாவாயைச் செலுத்துவான்.

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 8 Statistics and Probability Ex 8.1

10th Maths Exercise 8.1 Samacheer Kalvi Question 1.
Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Solution:
Range R = L – S.

L – Largest value,
S – Smallest value.

(i) 63, 89, 98, 125, 79, 108, 117, 68.
Here L = 125
S = 63
∴ R = L – S = 125 – 63 = 62

Exercise 8.1 Class 10 Samacheer Question 2.
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Answer:
Range = 36.8
Smallest value (S) = 13.4
Range = L – S
36.8 = L – 13.4
L = 36.8 + 13.4 = 50.2
Largest value = 50.2

Ex 8.1 Class 10 Samacheer Question 3.
Calculate the range of the following data.

Solution:
Here the largest value = 650
The smallest value =400
∴ Range = L – S = 650 – 400
= 250

10th new syllabus maths exercise 8.1 Question 4.
A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
Solution:

10th Maths Statistics And Probability Question 5.
Find the variance and standard deviation of the wages of 9 workers given below:
₹ 310, ₹ 290, ₹ 320, ₹ 280, ₹ 300, ₹ 290, ₹ 320, ₹ 310, ₹ 280.
Solution:

10th Maths Exercise 8.1 Question 6.
A wall clock strikes the bell once at 1 o’clock, 2 times at 2 o’clock, 3 times at 3 o’clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell make a day.
Solution:

10th Maths Ex 8.1 Question 7.
Find the standard deviation of the first 21 natural numbers.
Solution:

10th Maths Exercise 8.1 In Tamil Question 8.
If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Answer:
The standard deviation of the data = 4.5
Each data is decreased by 5
The new standard deviation = 4.5

10th Maths 8.1 Question 9.
If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Solution:
If the standard deviation of a data is 3.6, and each of the data is divided by 3 then the new standard deviation is also divided by 3.
∴ The new standard deviation = \(\frac{3.6}{3}\)
= 1.2
The new variance = (standard deviation)²
\(=\sigma^{2}=1.2^{2}=1.44\)

10th Maths Chapter 8 Exercise 8.1 Question 10.
The rainfall recorded in various places of five districts in a week are given below.

Find its standard deviation.
Solution:

10th Maths 8.1 Tamil Question 11.
In a study about viral fever, the number of people affected in a town were noted as

Find its standard deviation.
Solution:
Let the assumed mean A = 35, C = 10

10th Maths 8th Chapter Question 12.
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find its standard deviation.

Solution:
Assumed mean A = 30.5, C = 4

12th Maths Exercise 8.1 Samacheer Kalvi Question 13.
The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Solution:
Assumed mean A = 11, C = 1

Samacheer Kalvi Guru 10th Maths Question 14.
For a group of 100 candidates the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Solution:

Samacheer Kalvi.Guru 10th Question 15.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Solution:
>
The given 5 number are 2, 4, a, b, 10, 12, 14.
a, b are 6 and 8.

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All these concepts of Chapter 14 Transportation in Plants and Circulation in Animals are explained very conceptually by the subject teachers in Tamilnadu State Board Solutions PDF as per the prescribed Syllabus & guidelines. You can download Samacheer Kalvi 10th Science Book Solutions Chapter 14 Transportation in Plants and Circulation in Animals State Board Pdf for free from the available links. Go ahead and get Tamilnadu State Board Class 10th Science Solutions of Chapter 1 Transportation in Plants and Circulation in Animals.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals

Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 14th Science Chapter 14 Transportation in Plants and Circulation in Animals Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 14 easily after studying the Tamilnadu State Board Class 14th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 14 Transportation in Plants and Circulation in Animals solutions of Class 14th by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 10th Science Transportation in Plants and Circulation in Animals Textual Evaluation Solved

I. Choose the Correct Answer

Transportation In Plants And Circulation In Animals Question 1.
Active transport involves ______.
(a) movement of molecules from lower to higher concentration.
(b) expenditure of energy.
(c) it is an uphill task.
(d) all of the above.
Answer:
(a) movement of molecules from lower to higher concentration.

Transportation In Plants And Circulation In Animals Class 10 Question 2.
Water which is absorbed by roots is transported to aerial parts of the plant through:
(a) cortex
(b) epidermis
(c) phloem
(d) xylem
Answer:
(d) xylem

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

Transportation In Plants And Circulation In Animals Book Back Answers Question 3.
During transpiration, there is loss of ______.
(a) carbon dioxide
(b) oxygen
(c) water
(d) none of the above.
Answer:
(c) water

10th Science Samacheer Kalvi Question 4.
Root hairs are:
(a) cortical cell
(b) projection of epidermal cell
(c) unicellular
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Samacheer Kalvi 10th Science Book Back Answers Question 5.
Which of the following process requires energy?
(a) active transport
(b) diffusion
(c) osmosis
(d) all of them.
Answer:
(a) active transport

Samacheer Kalvi 10th Science Guide Question 6.
The wall of human heart is made of:
(a) Endocardium
(b) Epicardium
(c) Myocardium
(d) All of the above
Answer:
(d) All of the above

Samacheerkalvi.Guru Science Question 7.
Which is the sequence of correct blood flow ______.
(a) ventricle – atrium – vein – arteries
(b) atrium – ventricle – veins – arteries
(c) atrium – ventricle – arteries – vein
(d) ventricles – vein – atrium – arteries.
Answer:
(c) atrium – ventricle – arteries – vein

10th Samacheer Kalvi Science Question 8.
A patient with blood group 0 was injured in an accident and has blood loss. Which blood group the doctor should effectively use for transfusion in this condition?
(a) O group
(b) AB group
(c) A or B group
(d) all blood group
Answer:
(a) O group

Samacheer Kalvi 10th Science Book Solutions Question 9.
‘Heart of heart’ is called ______.
(a) SA node
(b) AV node
(c) Purkinje fibres
(d) Bundle of His.
Answer:
(a) SA node

Samacheer Kalvi 10th Science Question 10.
Which one of the following regarding blood composition is correct?
(a) Plasma – Blood + Lymphocyte
(b) Serum – Blood + Fibrinogen
(c) Lymph – Plasma + RBC + WBC
(d) Blood – Plasma + RBC + WBC + Platelets
Answer:
(d) Blood – Plasma + RBC + WBC + Platelets

II. Fill in the Blanks

Samacheer Kalvi 10th Science Solutions Question 1.
______ involves evaporative loss of water from aerial parts.
Answer:
Transpiration.

Class 10 Science Samacheer Kalvi Question 2.
Water enters the root cell through a ______ plasma membrane.
Answer:
Osmosis.

Science Class 10 Samacheer Kalvi Question 3.
Structures in roots that help to absorb water are ______.
Answer:
Root hairs.

Samacheer Kalvi 10 Science Question 4.
Normal blood pressure is ______.
Answer:
120 mm / 80 mm Hg.

Class 10 Science Solutions Samacheer Kalvi Question 5.
The normal human heartbeat rate is about _______ time per minute.
Answer:
72 – 75.

III. Match the following

Science Samacheer Kalvi Question 1.

1. Symplastic pathway	(a) Leaf
2. Transpiration	(b) Plasmodesmata
3. Osmosis	(c) Pressure in xylem
4. Root Pressure	(d) Pressure gradient

Answer:

1. (b) Plasmodesmata
2. (a) Leaf
3. (d) Pressure gradient
4. (b) Pressure in xylem.

Samacheer Kalvi Class 10 Science Solutions Question 2.

1. Leukaemia	(a) Thrombocytes
2. Platelets	(b) Phagocyte
3. Monocytes	(c) Decrease in leucocytes
4. Leucopenia	(d) Blood Cancer
5. AB blood group	(e) Allergic condition
6. O blood group	(f) Inflammation
7. Eosinophil	(g) Absence of antigen
8. Neutrophils	(h) Absence of antibody

Answer:

1. (d) Blood Cancer
2. (a) Thrombocytes
3. (b) Phagocyte
4. (c) Decrease in leucocytes
5. (h) Absence of antibody
6. (g) Absence of antigen
7. (e) Allergic condition
8. (f) Inflammation.

IV. State whether True or False. If false write the correct statement.

10th Std Science Solutions Samacheer Kalvi Question 1.
The phloem is responsible for the translocation of food.
Answer:
True.

Samacheer Kalvi 10th Standard Science Question 2.
Plants lose water by the process of transpiration.
Answer:
True.

10th Standard Samacheer Kalvi Science Question 3.
The form of sugar transported through the phloem is glucose.
Answer:
False.
Correct Statement: The form of sugar transported through the phloem is Sucrose.

Samacheer Kalvi 10th Science Book Answers Question 4.
In the apoplastic movement, the water travels through the cell membrane and enter the cell.
Answer:
False.
Correct Statement: In the apoplastic movement, the water travels through the intercellular spaces and walls of the cell.

Samacheer Kalvi Guru 10th Science Question 5.
When guard cells lose water the stoma opens.
Answer:
False.
Correct Statement: When guard cells lose water, the stoma closed.

Samacheer Kalvi Science Question 6.
initiation and stimulation of heartbeat take place by nerves.
Answer:
True.

Samacheer Kalvi 10th Science Solution Question 7.
All veins carry deoxygenated blood.
Answer:
False.
Correct Statement: All veins carry deoxygenated blood except the Pulmonary vein.

Science 10th Samacheer Kalvi Question 8.
WBC defend the body from bacterial and viral infections.
Answer:
True.

Question 9.
The closure of the mitral and tricuspid valves at the start of the ventricular systole produces the first sound ‘LUBB’.
Answer:
True.

V. Answer in a word or Sentence.

Question 1.
Name two-layered protective covering of the human heart.
Answer:
Pericardium.

Question 2.
What is the shape of RBC in human blood?
Answer:
RBC’s are bioconcave and disc shaped.

Question 3.
Why is the colour of the blood-red?
Answer:
The colour of the blood is red, due to the presence of respiratory pigment Haemoglobin.

Question 4.
Which kind of cells are found in the lymph?
Answer:
Cells found in the lymphatics are lymphocytes.

Question 5.
Name the heart valve associated with the major arteries leaving the ventricles.
Answer:
Semilunar Valves.

Question 6.
Mention the artery which supplies blood to the heart muscle.
Answer:
Heart muscle receive oxygenated blood from coronary arteries that orginate from the aortic arch.

VI. Short Answer Questions.

Question 1.
What causes the opening and closing of guard cells of stomata during transpiration?
Answer:
During transpiration, the movement of (Potassium) ions, in and out of the guard cells, causes the opening and closing of stomate. When the water moves inside the guard cells, causing them to swell up and become turgid, making the stomata open. When guard cells cause water to move out of the cell, make guard cells shrunk, and the stomata pore closes.

Question 2.
What is cohesion?
Answer:
The force of attraction between the water molecules is called cohesion.

Question 3.
Trace the pathway followed by water molecules from the time it enters a plant root to the time it escapes into the atmosphere from a leaf.
Answer:

Question 4.
What would happen to the leaves of a plant that transpires more water than its absorption in the roots?
Answer:
If the rate of transpiration exceeds the rate of absorption, the stomata get closed the cells lose their turgidity and the plant wilts.

Question 5.
Describe the structure and working of the human heart.
Answer:
The human heart is four-chambered. The two upper thin-walled chambers, Auricle or Atria and the lower thick-walled chambers ventricles. The chambers are separated by a septum. The two auricles are separated by an interatrial septum and the two ventricles are separated by the interventricular septum.

The right atrium receives deoxygenated blood from different parts of the body through superior vena cava, inferior vena cava and coronary sinus. Pulmonary veins bring oxygenated blood from the lungs to the left atrium. Both the right and the left auricles pump blood into the right and left ventricles respectively.

From the right ventricle, the pulmonary arteries supply deoxygenated blood to the lungs. From the left ventricle, the Aorta carries the Oxygenated blood to the various organs of the body. The coronary arteries supply blood to the heart. This process is repeated again and again.

Question 6.
Why is the circulation in man referred to as double circulation?
Answer:
In human, blood passes twice through the heart to supply once to the body.
Double circulation involves: (i) Systemic circulation, (ii) Pulmonary circulation.
(i) Systemic circulation: In systemic circulation, from the left ventricle blood is pumped into the aorta and to various parts of the body.
(ii) Pulmonary circulation: In pulmonary circulation, from right ventricle deoxygenated blood is pumped into pulmonary artery which carries blood to the lungs for oxygenation.

Question 7.
What are heart sounds? How are they produced?
Answer:
The rhythmic closure and opening of the valves cause the sound of the heart. When the closure of the tricuspid and bicuspid valves after the beginning of ventricular systole, the sound ‘LUBB’ is produced. When the closure of the semilunar valve at the end of ventricular systole, the sound ‘DUPP’ is produced.

Question 8.
What is the importance of valves in the heart?
Answer:
The valves in heart are muscular flap that regulates the flow of blood in a single direction and prevents back flow of blood.

Question 9.
Who discovered the Rh factor? Why was it named so?
Answer:
The Rh factor was discovered by Landsteiner and Wiener in 1940 in Rhesus monkey. So it was named the Rh factor.

Question 10.
How are arteries and veins structurally different from one another?
Answer:

Artery	Vein
1. Distributing vessel	1. Collecting vessel
2. Pink in colour	2. Red in colour
3. Deep location	3. Superficial in location
4. Blood flow with high pressure	4. Blood flow with low pressure
5. Wall of an artery is strong, thick and elastic	5. Wall of a vein is weak, thin and non-elastic
6. All arteries carry oxygenated blood except pulmonary arteries	6. All veins carry deoxygenated blood except pulmonary veins
7. Internal valves are absent	7. Internal valves are present

Question 11.
Why is the Sinoatrial node called the pacemaker of heart?
Answer:
Although impulse is produced by the entire neuro muscular pathway, the frequency of impulse generation is maximum in case of Sino atrial node in comparison to other parts of pathway. Hence it guides the rhythm of heart beat and is called the pacemaker of the heart.

Question 12.
Differentiate between systemic circulation and pulmonary circulation.
Answer:

Systemic Circulation	Pulmonary Circulation
1. Circulation of oxygenated blood from the left ventricle of the heart to various organs of the body.	1. Circulation starts in the right ventricle of the heart and reaches the lungs with deoxygenated blood.
2. Return of deoxygenated blood to the right atrium.	2. Pulmonary Artery collects the oxygenated blood from the lungs.
3. Aorta carries oxygenated blood to all the organs of the body.	3. The Oxygenated blood is supplied to the left atrium of the heart by the Pulmonary.

Question 13.
The complete events of the cardiac cycle last for 0.8 sec. What is the timing for each event?
Answer:

1. Auricular systole – Contraction of auricles = 0.1 sec
2. Ventricular systole – Contraction of ventricle = 0.3 sec
3. Ventricular diastole – Relaxation of ventricle = 0.4 sec

VII. Give reasons for the following Statements.

Question 1.
Minerals cannot be passively absorbed by the roots.
Answer:
The minerals cannot be passively absorbed by the roots because

• The minerals are present in the soil as charged particles (ion) and cannot move across the cell membrane.
• The concentration of minerals in the soil is lower than the concentration of minerals in the root.

Question 2.
Guard cells are responsible for opening and closing of Stomata.
Answer:
Opening and closing of stomata takes place due to changes in turgor of guard cell. The turgor changes in the guard cells are due to entry and exit of water into and out of the guard cells. During the day, water from the subsidiary cells enter the guard cells making it fully turgid causing the stomata to open. During night time, water from guard cells enters the subsidiary cells makes the guard cells flaccid causing the stomata closes.

Question 3.
The movement of substances in the phloem can be in any direction.
Answer:
The movement of substance in the Phloem can be in any direction because, the food to reach the plant parts like stem, leaves, flower, bud, seeds etc, the movement can be upwards or downwards, that is bidirectional.

Question 4.
Minerals in the plants are not lost when the leaf falls.
Answer:
Minerals are remobilised from older drying leaves to younger leaves. Elements like phosphorous, sulphur, nitrogen, potassium are easily mobilised, while elements like calcium are not remobilised. This can be seen in decidous leaves.

Question 5.
The walls of the right ventricle are thicker than the right auricles.
Answer:
The walls of the right ventricle are thicker than the right auricle because, the right ventricle has to pump out the blood with force to the Pulmonary trunk, which bifurcates to form the right and left Pulmonary Arteries.

Question 6.
Mature RBC in mammals do not have cell organelles.
Answer:
The RBCs are devoid of nucleus, mitochondria ribosome and endoplasmic reticulum. The absence of these organelles accommodates more haemoglobin thereby maximising the oxygen carrying capacity of the cell. Biconcave shape increase the surface area for oxygen binding, loss of mitochondria allow the RBC to transport all the oxygen to tissues and loss of endoplasmic reticulum allows more flexibility for RBC to move through the narrow capillaries.

VIII. Long Answer Questions.

Question 1.
How do plants absorb water? Explain.
Answer:
There are millions of root hairs on the tip of the root, which absorb water and minerals by diffusion. Diffusion takes place across cell membranes. Root hairs are a thin-walled, slender, extension of Epidermal cell, that increases the surface area of absorption. Active transport utilises energy to pump molecules against a concentration gradient. Active transport is carried out by membrane – bound proteins. These proteins use energy to carry substances across the cell membrane.

The cell wall of root hair is permeable and allows the water and minerals to enter. The cell membrane is semi – permeable. So it allows movement of water molecules from the region of higher concentration to the region of lower concentration. Once the water enters the root hairs, the concentration of water molecules in the root hair cells become more than that of Cortex. So the water from the root hair moves to the cortical cells by osmosis and then reaches the xylem.

Due to transpiration, the water is lost from the leaves and pressure is created at the top to pull more water from the xylem to the mesophyll cells, by the process of Transpiration pull. This extends up to the roots causing the roots to absorb more water from the soil to ensure the continuous flow of water from the roots to the leaves.

Question 2.
What is transpiration? Give the importance of transpiration.
Answer:
The loss of water from the aerial parts of plant in the form of vapours is called transpiration.
Importance of transpiration:

1. Creates transpirational pull for transport of water.
2. Supplies water for photosynthesis.
3. Transports minerals from soil to all parts of the plant.
4. Cools the surface of the leaves by evaporation.
5. Keeps the cells turgid; hence, maintains their shape.

Question 3.
Why are leucocytes classified as granulocytes and agranulocytes? Name each cell and mention its functions.
Answer:
White blood corpuscles are colourless. They are nucleated cells. They are found in the bone marrow, spleen, thymus and lymph nodes. They are grouped into two categories:
1. Granulocytes: They contain granules in their cytoplasm. Their nucleus is irregular or lobed. The granulocytes are of three types:

• Neutrophils: They are large in size and have a 2 – 7 lobed nucleus. Their numbers are increased during infection and inflammation.
• Eosinophils: It has a bilobed nucleus. Their numbers increases during conditions of allergy and parasitic infections. It brings about the detoxification of toxins.
• Basophils: Basophils have a lobed nucleus. They release chemicals during the process of inflammation.

2. Agranulocytes: Granules are not found in the cytoplasm of these cells. Thy is of two types:

• Lymphocytes: Lymphocytes produce antibodies during bacterial and viral infections.
• Monocytes: They are the largest of the leucocytes and are amoeboid in shape. They are phagocytic and can engulf bacteria.

Question 4.
Differentiate between systole and diastole. Explain the conduction of heart beat.
Answer:
systole:
The contraction of heart is called systole.
diastole:
The relaxation of heart is called diastole.
Conduction of heart beat : The heart in human is myogenic. The cardiac cells with fastest rhythm are called the pacemaker cells. These cells are located in the right sino-atrial node. The impulse from the sino-atrial node. spreads as a wave of contraction over the right and left atrial wall pushing the blood through the atrio ventricular valves into the ventricle. Two special cardiac muscles fibres originate from the auriculo ventricular node and are called the bundle of his which runs down into the interventricular septum and the fibres spread into the ventricles. These fibres are called the Purkinje.

Question 5.
Enumerate the functions of blood.
Answer:
Functions of blood:

• Transport of respiratory gases (Oxygen and CO₂).
• Transport of digested food materials to different body cells.
• Transport of hormones.
• Transport of nitrogenous excretory products like ammonia, urea and uric acid.
• It is involved in the protection of the body and defence against diseases.
• It acts as a buffer and also helps in the regulation of pH and body temperature.
• It maintains proper water balance in the body.

IX. Assertion and Reasoning Questions.

Direction: In each of the following questions a statement of assertion (A) is given and a corresponding statement of Reason (R) is given just below it. Mark the correct statement as:
(a) If both A and R are true and R is the correct explanation of A.
(b) If both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) Both A and R are false.

Question 1.
Assertion: RBC plays an important role in the transport of respiratory gases.
Reason: RBC does not have cell organelles and nucleus.
Answer:
(a) If both A and R are true and R is the correct explanation of A.

Question 2.
Assertion: Persons with AB blood group are called universal recipients because they can receive blood from all groups.
Reason: Antibodies are absent in persons with AB blood group.
Answer:
(a) If both A and R are true and R is the correct explanation of A.

X. Higher Order Thinking Skills (HOTS) Questions

Question 1.
When any dry plant material is kept in water, they swell up. Name and define the phenomenon involved in this change.
Answer:
The swelling up is due to Imbibition. Imbibition is a type of diffusion in which a solid absorbs water and gets swelled up. If Imbibition were not there, seedlings would not have been able to emerge out of the soil.

Question 2.
Why are the wails of the left ventricle thicker than the other chambers of the heart?
Answer:
The left ventricles have thick walls because the ventricle have to pump out blood with force away from the heart.

Question 3.
Doctors use a stethoscope to hear the sound of the heart. Why?
Answer:
The heart sound is heard by placing the stethoscope on the chest. It is a useful diagnostic tool to identify and localize the health problems and diagnose disease.

Question 4.
How does the pulmonary artery and pulmonary vein differ in their function when compared to a normal artery and vein?
Answer:

1. All arteries carry oxygenated blood except the pulmonary artery which carry deoxygenated blood to the lungs.
2. All veins carry deoxygenated blood except the pulmonary vein which carry oxygenated blood from the lungs to the heart.

Question 5.
Transpiration is a necessary evil in plants. Explain.
Answer:
Transpiration is a necessary evil in plants because it is inevitable but potentially harmful. Loss of water from the plant results wilting and cause the death of a plant if a condition of drought is experienced.
But transpiration is a great significance for the plant.

• Water is conducted, in most tall plants due to transpiration pull.
• Minerals dissolved in water are distributed throughout the plant body by Transpiration Stream.
• Evaporation of water from the cells of leaves has a cooling effect on plants.
• The wet surface of leaf cells allows gaseous exchange.

Textbook Activities Solved

Activity 1.
Demonstration of Osmosis: A thistle funnel whose mouth is covered with a semipermeable membrane is filled with sucrose solution. It is kept inverted in a beaker containing water. The water will diffuse across the membrane due to osmosis and raise the level of the solution in the funnel.
Answer:
A thistle funnel, whose mouth is covered with parchment paper, which acts as a semi-permeable membrane, is filled with sucrose solution. The funnel is kept inverted in a beaker, containing water. The water from the beaker enters or diffuse across the membrane from the region of water higher concentration to the water lower concentration (Sucrose Solution) through Semi – Permeable membrane, by osmosis. This causes the rise of the level of a solution in the funnel.

Activity 2.
Demonstration of Root Pressure: Choose a small soft – stemmed plant. Cut the stem horizontally near the base with a blade in the morning. You will see drops of solution oozing out of the cut stem due to root pressure.
Answer:
We will see drops of solution Oozing out of the cell stem. The Root Pressure, which pushes ^the solution up from the base.

Activity 3.
Determining Heart Rate?
Answer:
Materials:

1. Stopwatch or Stop clock.

Procedure:

1. Have your partner to find the pulse in your wrist and count your heartbeats for 15 seconds while you are seated. Calculate your resting heart rate in beats per minute.
2. Have your partner to count your heart beats for 15 seconds after you jog or run for 5 minutes. Calculate your heart rate in beats per minute.

Analyse:
(i) What causes your pulse?
(ii) What causes the change in your heartbeat rate in each situation? You can write the answer yourself.
Answer:
(i) Pulse, rhythmic dilation of an artery generated by the opening and closing of the aortic valve in the heart. A pulse can be felt by applying firm fingertip pressure to the skin at sites where the arteries travel near the skin’s surface; it is more evident when surrounding muscles are relaxed.
(ii) You can write the answer yourself.

Samacheer Kalvi 10th Science Transportation in Plants and Circulation in Animals Additional Questions Solved

I. Fill in the blanks.

Question 1.
______ is responsible for the movement of water up to the base of the stem.
Answer:
Root pressure.

Question 2.
The other name for Red blood corpuscles (RBC) is called ______.
Answer:
Erythrocytes.

Question 3.
The apices of the flaps of tricuspid valves are held in position by ______.
Answer:
Chordaetendinae.

Question 4.
The supply of blood to the heart muscles is called ______ circulation.
Answer:
Coronary.

Question 5.
Sinuses are the body cavities which are called ______.
Answer:
Haemocoel.

II. Write True or False for the following statements. Write the true statement for the false statement.

Question 1.
Root hairs are a thin-walled, slender extension of Parenchyma cells.
Answer:
False.
Correct Statement: Root hairs are a thin-walled, slender extension of epidermal cell.

Question 2.
The transpiration pull sucks the water column from the xylem tubes so that the water is able to rise up in the tallest plants. Answer:
True.

Question 3.
The force of attraction between the molecules of companion cells is called cohesion.
Answer:
False.
Correct Statement: The force of attraction between the molecules of water is called cohesion,

Question 4.
Blood platelets help in the clotting of blood, form clot at the site of injury and prevent blood loss.
Answer:
True.

III. Match the following.

Question 1.

1. Universal Recipient	(a) Left Atrio Ventricular Valve
2. Blood	(b) Thin and non-elastic cells
3. Thymocytes	(c) Blood group AB
4. Universal Donor	(d) Thick and elastic cells
5. Mitral	(e) Blood group ‘O’
6. Veins	(f) Blood platelets
7. Arteries	(g) Connective tissue

Answer:

1. (c) Blood group AB
2. (g) Connective tissue
3. (f) Blood platelets
4. (e) Blood group ‘O’
5. (a) Left Atrio Ventricular Valve
6. (b) Thin and non-elastic cells
7. (d) Thick and elastic cells

IV. Answer the following in a word or a Sentence.

Question 1.
What is Root Pressure?
Answer:
As ion from the Soil is actively transported into the Vascular tissue of the root, water moves along and increases the pressure inside the Xylem. This pressure is called root pressure.

Question 2.
What is the role of valves in heart?
Answer:
The valves are muscular flap that regulates the flow of blood in a single direction and prevent back flow of blood.

Question 3.
What is capillary action?
Answer:
Water or any liquid rises in a capillary tube because of physical forces. This phenomenon is called capillary action.

Question 4.
What is normal heart beat in man? How does it occur?
Answer:
The normal heart beat in man is about 72 – 75 times per minute. Rhythmic contraction and expansion of heart causes heart beat.

Question 5.
What are capillaries?
Answer:
Capillaries are narrow tubes formed by branching of arterioles which then unite to form the venules and veins. Capillaries are formed of a single layer of endothelial cells.

Question 6.
What is a pulse?
Answer:
The expansion of the artery every time, the blood is forced into it is called pulse. Normal pulse rate ranges from 70 – 90 per minute.

Question 7.
What is Cardiac Cycle?
Answer:
The sequence of events occurring from the beginning to the completion of one heartbeat is called the Cardiac Cycle.

Question 8.
What is hypertension?
Answer:
The condition, where a prolonged or constant elevation of blood pressure exist is called hypertension (High blood pressure)

Question 9.
What is Apoplast pathway?
Answer:
The apoplastic movement of water occurs through the intercellular spaces and the walls of the cells with gradient energy.

Question 10.
Why are membrane – bound proteins called pumps?
Answer:
Active transport is carried out by membrane-bound proteins. These proteins use energy to carry substances across the cell membrane. So they are called pumps.

Question 11.
When does plasmolysis occur?
Answer:
Plasmolysis occur when water moves out of the cell and resulting in the shrinkage of cell membrane away from the cell wall.

Question 12.
What are Lymph nodes?
Answer:
Lymph nodes are small oval or fan-shaped structures, located along the length of lymphatic vessels.

V. Answer the following briefly.

Question 1.
Define the following.

1. Diffusion
2. Osmosis
3. Plasmolysis

Answer:

1. Diffusion: The movement of molecules in liquid and solids from a region of higher concentration to a region of lower concentration, without the utilization of energy is called Diffusion.
2. Osmosis: Osmosis is the movement of solvent or water molecules, from the region of higher concentration to the region of lower concentration through a semi – permeable membrane.
3. Plasmolysis: Plasmolysis is a process, which occurs, when water moves out of the cell and resulting in the shrinkage of the cell membrane, away from the cell wall.

Question 2.
Explain cardiac cycle.
Answer:
The sequence of events occurring from the beginning to the completion of one heart beat is called cardiac cycle. During cardiac cycle blood flows ’ through the chambers of the heart in a specific direction. Each cardiac cycle lasts about 0.8 second. The events during a single cardiac cycle involves

1. Atrial systole: Contraction of auricles (0.1 sec)
2. Ventricular systole: Contraction of ventricles (0.3 sec)
3. Ventricular diastole: Relaxation of ventricles (0.4 sec)

Question 3.
Explain the two types of Circulatory System.
Answer:
The two types of Circulatory System in animals are

1. Open type: The blood is pumped by the heart into blood vessels, that open into blood spaces called sinuses, which are the body cavities called haemocoel. Capillary System is absent. Eg. Arthropods, Molluscs, Ascidians.
2. Clsed type: The blood flows in a complete circuit around the body, through specific blood vessels. The blood flows from arteries to veins, through small blood vessels called capillaries. Eg. Vertebrates.

Question 4.
What is Blood Pressure? How is blood pressure expressed?
Answer:
The force exerted during the flow of blood against the lateral walls of arteries is called Blood Pressure. Blood pressure is usually expressed in terms of systolic pressure and diastolic pressure.

1. Systolic pressure: The pressure, during ventricular systole, the left ventricle contracts and forces the blood into the aorta is called systolic pressure.
2. Diastolic pressure: The pressure, during diastole, the ventricle relaxes and the pressure falls to the lowest value, is called Diastolic pressure.

In a healthy adult, during the normal resting condition, the systolic and diastolic blood pressure is expressed as 120mm / 80mm Hg.

Question 5.
In a Tabular Column, mention the distribution of Antigen (RBC) and antibody (Plasma) in different blood groups.
Answer:

Blood Group	Antigens on RBC	Antibodies in Plasma	Can donate to	Can receive from
A	Antigen A	anti – b	A and AB	A and O
B	Antigen B	anti – a	B and AB	B and O
AB	Antigen A and B	No Antibody	AB	A, B, AB and O (Universal Recipient)
O	No Antigen	Both anti a and b	A, B, AB and O (Universal Donor)	O

Question 6.
Write a note on valves of human.
Answer:
Valves : The valves are the muscular flaps that regulate the flow of blood in a single direction and prevent back flow of blood. The heart contains three types of valves. Right atrioventricular valve : It is located between the right auricle and right ventricle.

It has three thin triangular leaf like flaps and therefore called tricuspid valve. The apices of the flaps are held in position by chordae tendinae arising from the muscular projection of the ventricle wall known as papillary muscles.

Left atrioventricular valve : It is located between the left auricle and left ventricle. It has two cusps and therefore called bicuspid or mitral valve.
Semilunar valves : The major arteries (pulmonary artery and aorta) which leave the heart have semilunar valves which prevent backward flow of blood into the ventricles. They are the pulmonary and aortic semilunar valves.

Question 7.
What is Sphygmo Manometer? Name the instruments used to measure Blood pressure.
Answer:
The sphygmomanometer is a clinical instrument used to measure blood pressure when a person is in a relaxed and resting condition. The pressure of the brachial artery is measured. It helps to estimate the state of blood circulation and the working of the heart. It helps to diagnose conditions such as increased or decreased blood pressure. Monometric and modem digital types are used to measure blood pressure.

Question 8.
What are the factors which affect transpiration?
Answer:
Transpiration is affected by several external and internal factors.

• External factors: Temperature, light, humidity, and wind speed.
• Internal factors: Number and distribution of Stomata, Percentage of open stomata, Water status of the plant and Canopy Structure.

VI. Draw a labelled diagram for the following.

Question 1.
(a) Root tip with root hairs.
(b) Guard cell in turgid and Flacid condition.
Answer:
(a)

(b)

Question 2.
Draw a labelled diagram of the internal structure of Human Heart.
Answer:

VII. Answer the following in detail.

Question 1.
Explain the types of Blood Circulation.
Answer:
The blood circulates in our body as oxygenated and deoxygenated blood. The types of circulation are:

1. Systemic Circulation: Circulation of oxygenated blood from the left ventricle of the heart to various organs of the body and return of deoxygenated blood to the right atrium. Aorta carries oxygenated blood to all the organs of the body.
2. Pulmonary Circulation: The path of pulmonary circulation starts in the right ventricle. Pulmonary artery arises from the right ventricle and reaches the lungs with deoxygenated blood. Pulmonary veins collect the oxygenated blood from the lungs and supplies it to the left atrium of the heart.
3. Coronary circulation: The supply of blood to the heart muscles (cardiac muscles) is called coronary circulation. Cardiac muscles receive oxygenated blood from coronary arteries, that originate from the aortic arch. Deoxygenated blood from the cardiac muscles, drains into the right atrium by the coronary sinuses.

In human, the blood circulates twice, through the heart in one complete cycle, called double circulation. The oxygenated blood does not mix with the deoxygenated blood.

Question 2.
Explain the Lymphatic System with diagram and mention the function of Lymph.
Answer:
The lymphatic system consists of lymphatic capillaries, lymphatic vessels, lymph nodes and lymphatic ducts.
Lymph is a fluid, that flows through the lymphatic system. The lymphatic capillaries unite to form large lymphatic vessels. Lymph nodes are small, oval or pear-shaped structures located along the lymphatic vessels.

Lymph from the intercellular spaces drains into lymphatic capillaries. Lymph is a colourless fluid formed when Plasma, Proteins and blood cells escape into intercellular spaces in the tissues, through the pores present in the walls of the capillaries. It is similar to blood plasma, but is colourless and contains fewer proteins. The lymph contains a very small amount of nutrients, Oxygen, CO₂, water and WBC.

Functions of Lymph:

• Supplies nutrients and oxygen to those parts where blood cannot reach.
• It drains away excess tissue fluid and metabolites and returns proteins to the blood from tissue spaces.
• The lymph also carries absorbed fats from small intestine to the blood. The lymphatic capillaries of internal villi (lacteals) absorb digested fats.
• Lymphocytes in the lymph defend the body from infections.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
What do you know about the following?

1. Anaemia
2. Leucocytosis
3. Leukopenia
4. Thrombocytopenia

Answer:

1. Anaemia → Decrease in the number of Erythrocytes.
2. Leucocytosis → Increase in the number of Leukocytes.
3. Leukopenia → Decrease in the number of Leukocytes.
4. Thrombocytopenia → Decrease in the number of Thrombocytes.

Question 2.
The cardiac pacemaker in a patient fails to function normally. The doctor finds that an artificial pacemaker is to be grafted in him. It is likely that it will be drafted at the site of.
Answer:
Sino-atrial node, which acts as the pacemaker of the heart because it is capable of initiating impulse which can stimulate the heart muscles to contraction.

Question 3.
What do you know about the following important terms, which we come across every day?

1. Heart attack
2. Cardiac arrest
3. Heart Failure.

Answer:

1. Heart attack: The blood flow to the heart is blocked. A blockage of blood flow to the heart muscle. The heart muscle affected from lack of blood supply.
2. Cardiac arrest: The heart stops beating and needs to be restarted. Cardiac arrest is an electrical problem, triggered by a disruption of the heart’s rhythm.
3. Heart Failure: Heart Failure occurs, when the heart muscle fails to pump as much as blood as the body needs. It may come to a person suddenly.

Question 4.
Removal of ring wood of tissue outside the vascular cambium from the tree trunk kills it because:
Answer:
Food does not travel down and root becomes starved (Translocation of food).

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules

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Samacheer Kalvi 10th Science Atoms and Molecules Textual Solved Problems

I. Calculation of molar mass:

Atoms And Molecules Class 10 Samacheer Question 1.
Calculate the gram molar mass of the following.
(i) H₂O
(ii) CO₂
(iii) Ca₃(PO₄)₂
Solution:
(i) H₂O
Atomic masses of H = 1, O = 16
Gram molar mass of H₂O = (1 × 2) + (16 × 1) = 2 + 16
Gram molar mass of H₂O = 18 g.

(ii) CO₂
Atomic masses of C = 12, O = 16
Gram molar mass of CO₂ = (12 × 1) + (16 × 2) = 12 + 32
Gram molar mass of CO₂ = 44 g.

(iii) Ca₃(PO₄)₂
Atomic masses of Ca = 40, P = 30, O = 16.
Gram molar mass of Ca₃(PO₄)₂ = (40 × 3) + [30 + (16 × 4)] × 2
= 120 + (94 × 2)
= 120 + 188
Gram molar mass of Ca₃(PO₄)₂ = 308 g.

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II. Calculation based on number of moles from mass and volume:

Atoms And Molecules Class 10 Pdf Question 1.
Calculate the number of moles in 46 g of sodium.
Solution:
Number of moles = \(\frac{\text { Mass of the element }}{\text { Atomic mass of the element }}\)
Atomic mass of the element = \(\frac { 46 }{ 23 }\) = 2 moles of sodium

10th Science Atoms And Molecules Question 2.
5.6 litre of Oxygen at S.T.P?
Solution:
Given volume of O₂ at,
Number of moles = \(\frac{\text { S.T.P. }}{\text { Molar volume at } \mathrm{S} . \mathrm{T} \cdot \mathrm{P}}\)
Molar volume at S.T.P = \(\frac { 46 }{ 23 }\) = 2 moles
Number of moles of oxygen = \(\frac{5.6}{22.4}\) = 0.25 mole of oxygen

10th Science Atoms And Molecules Book Back Answers Question 3.
Calculate the number of moles of a sample that contains 12.046 × 10²³ atoms of iron?
Solution:
Number of moles = \(\frac{\text { Number of atoms of iron }}{\text { Avogadro’s number }}\)
= 12.046 × 10²³ / 6.023 × 10²³
= 2 moles of iron.

III. Calculation of mass from a mole.

Atoms And Molecules Class 10 Book Back Answers Question 1.
0.3 mole of aluminium (Atomic mass of Al = 27).
Solution:
Number of moles = \(\frac{\text { Mass of Al }}{\text { Atomic mass of Al }}\)
Mass = No. of moles × atomic mass
So, mass of Al = 0.3 × 27 = 8.1 g.

Atoms And Molecules – Class 10 Samacheer Question 2.
2.24 litre of SO₂ gas at S.T.P?
Solution:
Molecular mass of SO₂ = 32 + (16 × 2) = 32 + 32 = 64
Number of moles of SO₂ = \(\frac{\text { Given volume of } \mathrm{SO}_{2} \text { at } \mathrm{S} . \mathrm{T.P}}{\text { Molar volume } \mathrm{SO}_{2} \text { at } \mathrm{S} . \mathrm{T.P}}\)
= \(\frac{2.24}{22.4}=0.1 \mathrm{mole}\)
Number of moles = \(\frac{\text { Mass }}{\text { Molecular mass }}\)
Mass = No. of moles × molecular mass
Mass = 0.1 × 64
Mass of SO₂ = 6.4 g.

10th Chemistry Atoms And Molecules Question 3.
1.51 × 10²³ molecules of water
Solution:
Molecular mass of H₂O = 18
Number of moles = \(\frac{\text { Number of molecules of water }}{\text { Avogadro’s number }}\)
= 1.51 × 10²³ / 6.023 × 10²³ = 1 / 4 = 0.25 mole
Number of moles = \(\frac{\text { Mass }}{\text { Molecular mass }}\)
0.25 = mass / 18
Mass = 0.25 × 18
Mass = 4.5 g.

Atoms And Molecules Class 10 Questions And Answers Question 4.
5 × 10²³ molecules of glucose?
Solution:
Molecular mass of glucose = 180
Mass of glucose = \(\frac{\text { Molecular mass } \times \text { number of particles }}{\text { Avogadro’s number }}\)
= (180 × 5 × 10²³) / 6.023 × 10²³
= 149.43 g.

IV. Calculation based on the number of atoms/molecules.

10th Science Atoms And Molecules Pdf Question 1.
Calculate the number of molecules in 11.2 litre of CO₂ at S.T.P
Solution:
Number of moles of CO₂ = \(\frac{\text { Volume at S.T.P }}{\text { Molar volume }}\)
= \(\frac { 11.2 }{ 22.4 }\)
= 0.5 mole.
Number of molecules of CO₂ = Number of moles of CO₂ × Avogadro’s number
= 0.5 × 6.023 × 10²³
= 3.011 × 10²³ molecules of CO₂.

Atoms And Molecules Class 10 Question 2.
Calculate the number of atoms present in 1 gram of gold (Atomic mass of Au = 198).
Solution:
Number of atoms of Au = \(\frac{\text { Mass of Au } \times \text { Avogadro’s number }}{\text { Atomic mass of Au }}\)
Atomic mass of Au = \(\frac{1}{198} \times 6.023 \times 10^{23}\)
Number of atoms of Au = 3.042 × 10²¹ g.

10th Atoms And Molecules Book Back Answers Question 3.
Calculate the number of molecules in 54 gm of H₂O
Solution:
Number of molecules = \(\frac{(\text { Avogadro number } \times \text { Given mass })}{\text { Gram molecular mass }}\)
Number of molecules of water = 6.023 × 10²³ × \(\frac { 54 }{ 18 }\)
= 18.069 × 10²³ molecules.

Atom And Molecules Class 10 Question 4.
Calculate the number of atoms of oxygen and carbon in 5 moles of CO₂.
Solution:

• 1 mole of CO₂ contains 2 moles of oxygen.
• 5 moles of CO₂ contains 10 moles of oxygen
  Number of atoms of oxygen = number of moles of oxygen × Avogadro’s number
  = 10 × 6.023 × 10²³ = 6.023 × 10²⁴ atoms of Oxygen.
• 1 mole of CO₂ contains 1 mole of carbon
• 5 moles of CO₂ contains 5 moles of carbon
  No. of atoms of carbon = No.of moles of carbon × Avogadro’s number
  = 5 × 6.023 × 10²³ = 3.011 × 10²⁴ atoms of Carbon.

V. Calculation based on molar volume

Calculate the volume occupied by:
Question 1.
2.5 mole of CO₂ at S.T.P.
Solution:
\(\begin{array}{l}{\text { Number of moles of } \mathrm{CO}_{2}=\frac{\text { Given volume at S.T.P }}{\text { Molar volume at S.T.P }}} \\ {\qquad 2.5 \text { mole of } \mathrm{CO}_{2}=\frac{\text { Volume of } \mathrm{CO}_{2} \text { at } \mathrm{S} . \mathrm{TP}}{22.4}}\end{array}\)
Volume of CO₂ at S.T.P = 22.4 × 2.5 = 56 litres.

Question 2.
3.011 × 10²³ of ammonia gas molecules?
Solution:
Number of moles = \(\frac{\text { Number of molecules }}{\text { Avogadro’s number }}\)
= 3.011 × 10²³ / 6.023 × 10²³
= 2 moles
Volume occupied by NH₃ = number of moles × molar volume
= 2 × 22.4 = 44.8 litres at S.T.P.

Question 3.
14 g nitrogen gas?
Solution:
Number of moles = \(\frac { 14 }{ 28 }\) = 0.5 mole
Volume occupied by N₂ at S.T.P = No. of moles × molar volume
= 0.5 × 22.4
= 11.2 litres.

VI. Calculation based on % composition.

Question 1.
Calculate % of S in H₂SO₄
Solution:
Molar mass of H₂SO₄ = (1 × 2) + (32 × 1) + (16 × 4)
= 2 + 32 + 64
= 98 g.
\(\begin{array}{l}{\% \text { of } \mathrm{S} \text { in } \mathrm{H}_{2} \mathrm{SO}_{4}=\frac{\text { Mass of sulphur }}{\text { Molar mass of } \mathrm{H}_{2} \mathrm{SO}_{4}} \times 100} \\ {\% \text { of } \mathrm{S} \text { in } \mathrm{H}_{2} \mathrm{SO}_{4}=\frac{32}{98} \times 100}\end{array}\)
= 32.65 %.

Samacheer Kalvi 10th Science Atoms and Molecules Textual Evaluation Solved

I. Choose the best answer.

Question 1.
Which of the following has the smallest mass?
(a) 6.023 × 10²³ atoms of He
(b) 1 atom of He
(c) 2 g of He
(d) 1 mole atoms of He.
Answer:
(b) 1 atom of He
Hint:
(a) 6.023 × 10²³ atoms of He = 1 mole
Mass of 1 mole of He = 4 g (or) 0.004 kg.

(b) Mass of 1 atom of He =?
Mass of 6.023 × 10²³ atoms of He = 0.004 kg.
Mass of 1 atom of He = \(\frac{0.004}{6.023 \times 10^{23}}=6.6423 \times 10^{-27} \mathrm{kg}\)

(c) 2 g of He = Mass = 0.002 kg.

(d) 1 mole atoms of He = 4 g = 0.004 kg.
So (b) is the smallest mass as 6.6423 × 10^-27 kg.

Question 2.
Which of the following is a triatomic molecule?
(a) Glucose
(b) Helium
(c) Carbon dioxide
(d) Hydrogen
Answer:
(c) Carbon dioxide

Question 3.
The volume occupied by 4.4 g of CO₂ at S.T.P _____.
(a) 22.4 litre
(b) 2.24 litre
(c) 0.24 litre
(d) 0.1 litre.
Answer:
(b) 2.24 litre
Hint:
Molar volume of CO₂ = 22.4 litre.
The volume occupied by 1 mole.
i.e. 44 g (molar mass) of CO₂.
44 g of CO₂ occupied 22.4 litre of volume.
4.4 g of CO₂ will occupy \(\frac{22.4}{44}\) × 4.4 = \(\frac{22.4}{10}\) = 2.24 litre.
So, answer (b) is correct.

Question 4.
Mass of 1 mole of Nitrogen atom is:
(a) 28 amu
(b) 14 amu
(c) 28 g
(d) 14 g
Answer:
(c) 28 g

Question 5.
Which of the following represents 1 amu?
(a) Mass of a C – 12 atom
(b) Mass of a hydrogen atom
(c) \(\frac { 1 }{ 2 }\)th of the mass of a C – 12 atom
(d) Mass of O – 16 atom
Answer:
(c) 1 / 12th of the mass of a C – 12 atom
Hint: By definition 1 amu is defined as precisely 1 / 12th the mass of an atom of carbon – 12.
So, answer (c) is correct.

Question 6.
Which of the following statement is incorrect?
(a) One gram of C – 12 contains Avogadro’s number of atoms.
(b) One mole of oxygen gas contains Avogadro’s number of molecules.
(c) One mole of hydrogen gas contains Avogadro’s number of atoms.
(d) One mole of electrons stands for 6.023 × 10²³ electrons.
Answer:
(a) One gram of C – 12 contains Avogadro’s number of atoms.

Question 7.
The volume occupied by 1 mole of a diatomic gas at S.T.P is _____.
(a) 11.2 litre
(b) 5.6 litre
(c) 22.4 litre
(d) 44.8 litre.
Answer:
(c) 22.4 litre
Hint: By definition 1 mole of any gas at S.T.P occupies molar volume i.e. 22.4 litres.
So (c) is the correct answer.

Question 8.
In the nucleus of \(_{20} \mathrm{Ca}^{40}\), there are _____.
(a) 20 protons and 40 neutrons
(b) 20 protons and 20 neutrons
(c) 20 protons and 40 electrons
(d) 40 protons and 20 electrons.
Answer:
(b) 20 protons and 20 neutrons
Hint:
\(_{20} \mathrm{Ca}^{40}\)
20 = Atomic number = Number of protons (or) Number of electrons
40 = Mass number = Number of protons + Number of neutrons
\(_{20} \mathrm{Ca}^{40}\) contains 20 protons, 20 electrons and 20 neutrons.
So the answer (b) is correct.

Question 9.
The gram molecular mass of oxygen molecule is:
(a) 16 g
(b) 18 g
(c) 32 g
(d) 17 g
Answer:
(b) 18 g

Question 10.
1 mole of any substance contains ______ molecules.
(a) 6.023 × 10²³
(b) 6.023 × 10^-23
(c) 3.0115 × 10²³
(d) 12.046 × 10²³.
Answer:
(a) 6.023 × 10²³
Hint:
Avogadro’s law states that 1 mole of any substance contains 6.023 × 10²³ molecules.
So the answer (a) is correct.

II. Fill in the blanks

Question 1.
Atoms of different elements having ______ mass number, but ______ atomic numbers are called isobars.
Answer:
Same, different.

Question 2.
Atoms of different elements having the same number of _____ are called isotones.
Answer:
Neutrons.

Question 3.
Atoms of one element can be transmuted into atoms of other elements by _____.
Answer:
Artificial transmutation.

Question 4.
The sum of the numbers of protons and neutrons of an atom is called its _____.
Answer:
Mass number.

Question 5.
Relative atomic mass is otherwise known as _____.
Answer:
Standard atomic weight.

Question 6.
The average atomic mass of hydrogen is _____ amu.
Answer:
1.008 amu.

Question 7.
If a molecule is made of similar kind of atoms, then it is called ______ atomic molecule.
Answer:
Homo.

Question 8.
The number of atoms present in a molecule is called its _____.
Answer:
Atomicity.

Question 9.
One mole of any gas occupies _____ ml at S.T.P.
Answer:
22400.

Question 10.
Atomicity of phosphorous is _____.
Answer:
4.

III. Match the following.

Question 1.

a. 8 g of O2	i. 4 moles
b. 4 g of H2	ii. 0.25 moles
c. 52 g of He	iii. 2 moles
d. 112 g of N2	iv. 0.5 moles
e. 35.5 g of Cl2	v. 13 moles

Answer:
a – ii, b – iii, c – v, d – i, e – iv.

IV. True or False: (If false give the correct statement)

Question 1.
Two elements sometimes can form more than one compound.
Answer:
True.

Question 2.
Noble gases are Diatomic
Answer:
False.
Correct Statement: Noble gases are monoatomic

Question 3.
The gram atomic mass of an element has no unit?
Answer:
False.
Correct Statement: The gram atomic mass of an element is expressed in the unit grams.

Question 4.
1 mole of Gold and Silver contain the same number of atoms?
Answer:
True

Question 5.
The molar mass of CO₂ is 42 g?
Answer:
False.
Correct Statement: The molar mass of CO₂ is (12 + 32) = 44 g.

V. Assertion and Reason:

Answer the following questions using the data given below:
(i) A and R are correct, R explains the A.
(ii) A is correct, R is wrong.
(iii) A is wrong, R is correct.
(iv) A and R are correct, R doesn’t explain A.

Question 1.
Assertion: Atomic mass of aluminium is 27
Reason: An atom of aluminium is 27 times heavier than 1 / 12th of the mass of the C – 12 atoms.
Answer:
(i) A and R are correct, R explains the A.

Question 2.
Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.
Reason: The natural abundance of Chlorine isotopes are not equal.
Answer:
(i) A and R are correct, R explains the A.

VI. Short Answer Questions

Question 1.
Define Relative atomic mass.
Answer:
Relative atomic mass of an element is the ratio between the average mass of its isotopes to 1 / 12th part of the mass of a carbon – 12 atom. It is denoted as A_r.
\(\text { Relative atomic mass }=\frac{\text { Average mass of the isotopes of the element }}{1 / 12^{\text {th }} \text { of the mass of one Carbon- } 12 \text { atom }}\).

Question 2.
Write the different types of isotopes of oxygen and its percentage abundance.
Answer:
Isotopes of oxygen:

The atomic mass of oxygen = (15.9949 × 0.99757) + (16.9991 × 0.00038) + (17.9992 × 0.00205) = 15.999 amu.

Question 3.
Define Atomicity.
Answer:
The number of atoms present in the molecule is called atomicity.

Question 4.
Give any two examples for heteroatomic molecules.
Answer:
Heterodiatomic molecules.
e.g.,

• HCl
• NaCl.

Question 5.
What is Molar volume of a gas?
Answer:
The volume occupied by one mole of any gas at STP is called molar volume. Its value is equal to 22.4 litre or 22400 ml or 22400 cm³ or 2.24 × 10^-2 m³.

Question 6.
Find the percentage of nitrogen in ammonia.
Answer:
Ammonia – NH₃ = Molar mass = 14 + 3 = 17
Mass % of Nitrogen = \(\frac{14}{17} \times 100\) = 82.35%.

VII. Long Answer Questions.

Question 1.
Calculate the number of water molecule present in one drop of water which weighs 0.18 g.
Answer:
One mole of water weighs 18 g.
18 g of water contains 6.023 × 10²³ water molecules.
∴ 0.18 g of water contains
= \(\frac{6.023×10^{23}}{18}\) × 0.18
= 6.023 × 10²¹ water molecules

Question 2.
N₂ + 3H₂ → 2NH₃ (The atomic mass of nitrogen is 14, and that of hydrogen is 1)

• 1 mole of nitrogen (____ g) + ____.
• 3 moles of hydrogen (____ g) → ____.
• 2 moles of ammonia (____ g).

Answer:
N₂ + 3H₂ → 2NH₃

• 1 mole of N₂ = 28 g
• 3 moles of H₂ = 6 g
• 2 moles of NH₃ = 34 g
  ⇒ 1 mole of nitrogen (28 g) + 3 moles of hydrogen (6 g) → 2 moles of Ammonia (34 g).

Question 3.
Calculate the number of moles in:
(i) 27 g of Al
(ii) 1.51 × 10²³ molecules of NH₄Cl
Answer:
(i) 27 g of Al
Number of moles = \(\frac{\text { Mass }}{\text { Atomic mass }}\)
Number of moles in 27 g of Al = \(\frac{27}{27}\) = 1 mole.

(ii) 1.51 × 10²³ molecules of NH₄Cl
Number of moles = \(\frac{\text { Number of molecules }}{\text { Avogadro’s number }}=\frac{1.51 \times 10^{23}}{6.023 \times 10^{23}}=0.25 \text { moles. }\).

Question 4.
Give the salient features of ‘Modern atomic theory’.
Answer:
(i) An atom is no longer indivisible.
(ii) Atoms of the same element may have different atomic mass.
Eg: isotopes ₁₇Cl³⁵, ₁₇Cl³⁷.
(ii) Atoms of different elements may have same atomic masses.
Eg: Isobars ₁₈Ar⁴⁰, ₂₀Ca⁴⁰.
(iv) Atoms of one element can be transmuted into atoms of other elements. An atom is no longer indestructible.
(v) Atoms may not always combine in a simple whole number ratio.
Eg: Glucose C₆H₁₂O₆ C : H : O = 6 : 12 : 6 or 1 : 2 : 1.
(vi) Atom is the smallest particle that takes part in a chemical reaction.
(vii) The mass of an atom can be converted into energy (E = mc²).

Question 5.
Derive the relationship between Relative molecular mass and Vapour density.
Answer:
(i) The Relative Molecular Mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of Hydrogen.

(ii) Vapour density is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
Vapour Density (V.D.) = \(\frac{\text { Mass of a given volume of gas or vapour at S.T.P }}{\text { Mass of same volume of hydrogen }}\).

(iii) According to Avogadro’s law, equal volumes of all gases contain equal number of molecules. Thus, let the number of molecules in one volume = n, then

(iv) V.D at STP = \(\frac{\text { Mass of “n’molecules of a gas or vapour at S.T.P }}{\text { Mass of ‘n’molecules of hydrogen }}\)
Cancelling ‘n’ which is common, you get
V.D = \(\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P. }}{\text { Mass of } 1 \text { molecules of hydrogen }}\).

(v) Since hydrogen is diatomic
\(\mathrm{V} . \mathrm{D} .=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P. }}{\text { Mass of } 2 \text { atoms of hydrogen }}\).

(vi) By comparing the definition of relative molecular mass and vapour density we can write as follows.
\(\mathrm{V.D.}=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P. }}{2 \times \text { Mass of } 1 \text { atom of hydrogen }}\)
Relative molecular mass (hydrogen scale) \(=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at STP. }}{\text { Mass of } 1 \text { atom of hydrogen }}\).

(vii) By substituting the relative molecular mass value in vapour density definition, we get
Vapour density (V.D.) = Relative molecular mass / 2
⇒ 2 × vapour density = Relative molecular mass of a gas.

VIII. HOT Questions

Question 1.

Calcium carbonate is decomposed on heating in the following reaction CaCO₃ → CaO + CO₂
(i) How many moles of Calcium carbonate are involved in this reaction?
Answer:
One mole

(ii) Calculate the gram molecular mass of calcium carbonate involved in this reaction.
Answer:
Gram molecular mass of CaCO₃
= 40 + 12 + 3(16)
= 100 g

(iii) How many moles of CO₂ are there in this equation?
Answer:
One mole.
IX. Solve the following problems.

Question 1.
How many grams are there in the following?
Answer:
Formula = No. of moles (n) × (Gram molecular mass)

(i) 2 moles of hydrogen molecule, H₂
Answer:
Mass of 2 moles of H₂ molecule
= 2 × 2 = 4 g

(ii) 3 moles of chlorine molecule, Cl₂
Answer:
Gram molecular mass of 3 moles of Cl₂
= 3 × 71 = 213 g

(iii) 5 moles of sulphur molecule, S₂
Answer:
Gram molecular mass of 5 moles of S₂
= 5 × 8(32)
= 5 × 256 = 1280 g

(iv) 4 moles of phosphorous molecule, P₄
Answer:
Gram molecular mass of 4 moles of P₂
= 4 × 4(31)
= 4 × 124 = 496 g

Question 2.
Calculate the % of each element in calcium carbonate. (Atomic mass: C – 12, O – 16, Ca – 40)
Solution:
Calcium carbonate: CaCO₃
Molar mass of CaCO₃ = 40 + 12 + (16 × 3) = 100 g
% of Calcium \(=\frac{40}{100} \times 100=40 \%\)
% of Carbon \(=\frac{12}{100} \times 100=12 \%\)
% of Oxygen \(=\frac{48}{100} \times 100=48 \%\).

Question 3.
Calculate the % of oxygen in Al₂(SO₄)₃. (Atomic mass: Al – 12, O – 16, S – 32)
Solution:
Aluminium Sulphate – Al₂(SO₄)₃
Molar mass of Aluminium Sulphate = (27 × 2) + (32 × 3) + (16 × 12) = 54 + 96 + 192 = 342 g
% of Oxygen \(=\frac{192}{342} \times 100=56.14 \%\).

Question 4.
Calculate the % relative abundance of B – 10 and B – 11, if its average atomic mass is 10.804 amu.
Solution:
The average atomic mass of Boron = 10.804 amu.
% relative abundance of B – 10 = ?
% relative abundance of B – 11 = ?
Let the fraction of relative abundance of B – 10 = x
Let the fraction of relative abundance of B – 11 = y
x + y = 1
y = 1 –  x
Relative abundance = x (10) + (1 – x) (11) = 10.804 amu
⇒ 10x + 11 – 11x = 10.804 amu
⇒ 11 – x = 10.804 amu
⇒ -x = 10.804 – 11
⇒ -x = -0.196
⇒ x = 0.196
x = % abundance of B – 10 = 0.196 × 100 = 19.6 %
y = % abundance of B – 11 = 100 – 19.6 = 80.4 %
Percentage abundance of B – 10 = 19.6 %
Percentage abundance of B – 11 = 80.4 %.

Activities

Question 1.
Complete the following table by filling the appropriate values / terms

Solution:

Question 2.
Classify the following molecules based on their atomicity and fill in the table:
Fluorine (F₂), Carbon dioxide (CO₂), Phosphorous (P₄), Sulphur (S₈), Ammonia (NH₃), Hydrogen iodide (HI), Sulphuric Acid (H₂SO₄), Methane (CH₄), Glucose (C₆H₁₂O₆), Carbon monoxide (CO)

Solution:

Question 3.
Under same conditions of temperature and pressure if you collect 3 litres of O₂, 5 litres of Cl₂ and 6 litres of H₂,

1. Which has the highest number of molecules?
2. Which has the lowest number of molecules?

Solution:
Number of moles of O₂ \(=\frac{\text { Volume at S.T.P }}{\text { Molar volume }}\) \(=\frac{3}{22.4}\) = 0.1339 moles
Number of molecules = Number of moles × Avogadro number
= 0.1339 × 6.023 × 10²³
= 0.8064 × 10²³
= 8.064 × 10²² O₂ molecules.
Number of moles of Cl₂ = \(\frac{5}{22.4}\) = 0.2232 moles
Number of molecules = 0.2232 × 6.023 × 10²³ = 1.344 × 10²³ molecules.
Number of moles of H₂ = \(\frac{6}{22.4}\) = 0.2678 moles
Number of molecules = 0.2678 × 6.023 × 10²³ = 1.6129 × 10²³ molecules.

1. 6 litres of H₂ has the highest number of molecules.
2. 3 litres of O₂ has the lowest number of molecules.

Samacheer Kalvi 10th Science Atoms and Molecules Additional Question Solved

I. Choose the best answer.

Question 1.
Which of the following pair indicates isotopes?
(a) \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\)
(b) \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\)
(c) \(_{6} \mathrm{C}^{13},_{7} \mathrm{N}^{14}\)
(d) \(_{33} \mathrm{As}^{77},_{34} \mathrm{Se}^{78}\).
Answer:
(a) \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\)

Question 2.
The mass of a proton is equal to:
(a) 1 amu
(b) \(\frac{1}{12^{th}}\) of the mass of a C – 12 atom
(c) zero
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 3.
The sum of the number of protons and neutrons of an atom is called its _____.
(a) nucleus
(b) atomic number
(c) mass number
(d) relative atomic mass.
Answer:
(c) mass number

Question 4.
Total number of electrons present in 1.7 g of NH₃ is:
(a) 6.023 × 10²³
(b) 6.023 × 10²⁴
(c) 6.023 × 10²²
(d) 6.023 × 10²⁵
Answer:
(a) 6.023 × 10²³

Question 5.
An isotope of hydrogen without neutrons is _____.
(a) Deuterium \(_{1} \mathrm{H}^{2}\)
(b) Protium \(_{1} \mathrm{H}^{1}\)
(c) Tritium \(_{1} \mathrm{T}^{3}\)
(d) Heavy hydrogen \(_{1} \mathrm{D}^{2}\).
Answer:
(b) Protium \(_{1} \mathrm{H}^{1}\)

Question 6.
The isotope tritium contains 1 proton and neutron in the nucleus.
(a) 1
(b) 2
(c) 3
(d) none
Answer:
(b) 2

Question 7.
Which one of the following element is used as the standard for measuring the relative atomic mass of an element in now a days?
(a) \(_{1} \mathrm{H}^{2}\)
(b) \(6^{\mathrm{O}^{12}}\)
(c) C – 12
(d) C – 14.
Answer:
(c) C – 12

Question 8.
The atom with no neutrons in the nucleus is:
(a) He
(b) Deuterium
(c) Tritium
(d) Protium
Answer:
(d) Protium

Question 9.
The average atomic mass of carbon is _____.
(a) 12 amu
(b) 12.84 amu
(c) 24.011 amu
(d) 12.011 amu.
Answer:
(d) 12.011 amu.

Question 10.
Which one of the following is the most abundant element in both the Earth’s crust and in the human body?
(a) Carbon
(b) Silicon
(c) Oxygen
(d) Hydrogen.
Answer:
(c) Oxygen

Question 11.
Gram molecular mass of H₂SO₄ is:
(a) 49 g
(b) 54 g
(c) 98 g
(d) 100 g
Answer:
(c) 98 g

Question 12.
Boron – 10 and Boron – 11 are called _____.
(а) isotopes
(b) isobars
(c) isotones
(d) isomers.
Answer:
(c) isotopes

Question 13.
Which of the following are found in the elementary state in nature?
(a) Hydrogen chloride
(b) Carbon dioxide
(c) Noble gases
(d) Oxygen.
Answer:
(c) Noble gases

Question 14.
Ammonia gas is formed by the following reaction
N₂(g) + 3H₂(g) → 2NH₃(g)
The volume of H₂ required to form 6 dm³ of NH₃ is:
(a) 9 dm³
(b) 10 dm³
(c) 4 dm³
(d) 2 dm³
Answer:
(a) 9 dm³

Question 15.
Which one of the following is a hetero diatomic molecule?
(a) O₂
(6) N₂
(c) HI
(d) CH₄.
Answer:
(c) HI

Question 16.
Which one of the following is a hetero triatomic molecule?
(a) H₂O
(b) BCl₃
(c) CH₄
(d) PCl₅.
Answer:
(a) H₂O

Question 17.
1 gm atom of nitrogen represents:
(a) 6.023 × 10² N₂ molecules
(b) 22.4 litre of N₂ at STP
(c) 11.2 L of N₂ at STP
(d) 28 g of nitrogen
Answer:
(c) 11.2 L of N₂ at STP

Question 18.
Find out the hetero diatomic molecule?
(a) Hydrogen
(b) Hydrogen chloride
(c) Methane
(d) Ammonia.
Answer:
(b) Hydrogen chloride

Question 19.
Which one of the following is an example of a polyatomic molecule?
(a) Sulphur
(b) Gold
(c) Sodium
(d) Helium.
Answer:
(a) Sulphur

Question 20.
The gram molar mass of CO₂ is:
(a) 44 g
(b) 100 g
(c) 4.4 g
(d) 22 g
Answer:
(a) 44 g

Question 21.
Which one of the following is an example of a polyatomic molecule?
(a) Fluorine
(b) Glucose
(c) Oxygen
(d) Sodium.
Answer:
(b) Glucose (C₆H₁₂O₆)

Question 22.
The gram molecular mass of water is _____.
(a) 18 amu
(b) 18 g
(c) 18 u
(d) 18.
Answer:
(b) 18 g

Question 23.
The value of Avogadro’s number is _____.
(a) 6.023 × 10^-23
(b) 6.023 × 10²³
(c) 22.4
(d) 22400.
Answer:
(b) 6.023 × 10²³

Question 24.
The value of molar volume is _____.
(a) 22.4 ml
(b) 22.4 litres
(c) 22400 litres
(d) 2.24 litres.
Answer:
(b) 22.4 litres

Question 25.
Which one of the following represent Avogadro’s law?
(a) V ∝ \(\frac{1}{n}\)
(b) V ∝ n
(c) V ∝ \(\frac{1}{n^{2}}\)
(d) V² ∝ \(\frac{1}{n}\).
Answer:
(b) V ∝ n

Question 26.
Which of the following has the highest number of molecules?
(a) 1 litre of N₂
(b) 2 litres of oxygen
(c) 5 litres of Cl₂
(d) 6 litres of Hydrogen.
Answer:
(d) 6 litres of Hydrogen.

Question 27.
Which one of the following has the lowest number of molecules?
(a) 1 litre of N₂
(b) 2 litres of H₂
(c) 3 litres of O₂
(d) 4 litres of Cl₂
Answer:
(a) 1 litre of N₂

Question 28.
2 × Vapour density is equal to _____.
(a) atomic mass
(b) valency
(c) relative molecular mass
(d) atomic number.
Answer:
(c) relative molecular mass

Question 29.
The value of gram molar mass of CO₂ is _____.
(a) 44 amu
(b) 44 g
(c) 44
(d) 44 kg.
Answer:
(b) 44 g
Hint: Molar mass = 12 + (16 × 2) = 44 g.

Question 30.
The number of moles of a sample that contain 36 g of water is _____.
(a) 1 mole
(b) 0.5 mole
(c) 4 moles
(d) 2 moles.
Answer:
(d) 2 moles
Hint: 18 g of water = 1 mole
36 g of water = \(\frac{1}{18} \times 36\) = 2 moles

Question 31.
Which of the following has the largest number of particles?
(a) 8 g of CH₄
(b) 4.4 g of CO₂
(c) 34.2 g of C₁₂H₂₂O₁₁
(d) 2 g of H₂.
Answer:
(d) 2 g of H_2.
Hint. 2 g = Molar mass = 1 mole = 6.023 × 10²³ particles.
Others are less.

Question 32.
The number of molecules in 16.0 g of oxygen is _____.
(a) 6.023 × 10²³
(b) 6.023 × 10^-23
(c) 3.01 × 10^-23
(d) 3.011 × 10²³
Answer:
(d) 3.011 × 10²³
Hint: 32 g of oxygen contain 6.023 × 10²³ molecules.
16 g of oxygen will contain
\(\frac{6.023 \times 10^{23}}{32} \times 16=3.011 \times 10^{23}\)

Question 33.
The percentage of hydrogen in H₂O is _____.
(a) 8.88
(b) 11.2
(c) 20.60
(d) 80.0.
Answer:
(b) 11.2
Hint: 1 mole of H₂O has 2.016 g of H₂
Percentage of H₂ = \(\frac{2.016}{18}\) × 100 = 11.2

Question 34.
Which of the following contains the largest number of molecules?
(a) 0.2 mole of H₂
(b) 8.0 g of H₂
(c) 17 g of H₂O
(d) 6.0 g of CO₂
Answer:
(b) 8.0 g of H₂
Hint: No. of moles = \(\frac{8}{2}\) = 4 moles.
No. of molecules = mole × Avogadro number = 4 × 6.023 × 10²³ = 2.409 × 10²⁴

Question 35.
One gram of which of the following contains the largest number of oxygen atoms?
(a) O
(b) O₂
(c) O₃
(d) All contain the same
Answer:
(c) O₃

Question 36.
The percentage by weight of O₂ in CaSO₄. (O = 16, Ca = 40, S = 32) is _____.
(a) 64 %
(b) 28.2 %
(c) 47.05 %
(d) 16.2 %.
Answer:
(c) 47.05 %
Hint: % by weight of O₂ = \(\frac{64}{136}\) × 100 = 47.05 %.

Question 37.
One mole of a gas occupies a volume of 22.4 L. This is derived from _____.
(a) Berzilliu’s hypothesis
(b) Gay – Lussac’s law
(c) Avogadro’s law
(d) Dalton’s law.
Answer:
(c) Avogadro’s law

Question 38.
Volume of gas at STP is 1.12 × 10^-7 cc. Calculate the number of molecules in it.
(a) 3.011 × 10²⁰
(b) 3.011 × 10¹²
(c) 3.011 × 10²³
(d) 3.011 × 10²⁴
Answer:
(b) 3.011 × 10¹²
Hint. 2.24 × 10^-3 c molecules 6.023 × 10²³ molecules
1.12 × 10-7 cc contains = \(\frac{6.023 \times 10^{23}}{22400} \times 1.12 \times 10^{-7}\)
= 3.011 × 10¹².

Question 39.
The number of molecules of CO₂ present in 44 g of CO₂ is _____.
(a) 6.023 × 10²³
(b) 3.011 × 10²³
(c) 12 × 10²³
(d) 3 × 10¹⁰.
Answer:
(a) 6.023 × 10²³ (Avogadro number).

Question 40.
The volume occupied by 4.4 g of CO₂ at S.T.P is _____.
(a) 22.4 L
(b) 2.24 L
(c) 0.224 L
(d) 0.1 L.
Answer:
(b) 2.24 L
Hint. 44 g of CO₂ at S.T.P occupies 22.4 L
4.4 g of CO₂ at S.T.P will occupy \(\frac{22.4}{44}\) × 4.4 = 2.24 L.

Question 41.
How many molecules at present in one gram of hydrogen?
(a) 6.023 × 10²³
(b) 3.011 × 10²³
(c) 2.5 × 10²³
(d) 1.5 × 10²³
Answer:
(b) 3.011 × 10²³
Hint: H₂ = Molar mass = 2 g
2 g of H₂ contains 6.023 × 10²³ molecules
∴ 1 g of H₂ will contain = \(\frac{6.023 \times 10^{23}}{2} \times 1\)
= 3.011 × 10²³ molecules.

Question 42.
Atoms which have the same number of protons but different number of neutrons are called _____.
(a) isotopes
(b) isomers
(c) allotropes
(d) isotones.
Answer:
(a) isotopes

Question 43.
Number of atoms which a molecule to sulphur contains is _____.
(a) 3
(b) 8
(c) 4
(d) 2.
Answer:
(b) 8 (S₈)

Question 44.
An example of a triatomic molecule is _____.
(a) Ozone
(b) Nitrogen
(c) Hydrogen
(d) Ammonia.
Answer:
(a) Ozone

Question 45.
The atomic mass of sodium is 23. The number of moles in 46 g of sodium is _____.
(a) 0.5
(b) 2
(c) 1
(d) 0.25.
Answer:
(b) 2
Hint:
No. of moles = \(\frac{\text { Mass }}{\text { Atomic mass }}=\frac{46}{23}\) = 2.

Question 46.
The number of atoms in a molecule of the elementary substance is called _____.
(a) Atomic number
(b) Avogadro number
(c) Atomic mass
(d) Atomicity.
Answer:
(d) Atomicity.

Question 47.
Avogadro number represents the number of atoms in _____.
(a) 12 g of C – 12
(b) 4.4 g of CO₂
(c) 320 g of Sulphur
(d) 1 g of C – 12
Answer:
(a) 12 g of C – 12

Question 48.
The number of moles in 5 grams of Calcium is _____.
(a) 0.5 mole
(b) 0.125 mole
(c) 1.25 mole
(d) 12.5 mole.
Answer:
(a) 0.125 mole
Hint:
No. of moles = \(\frac{\text { Mass }}{\text { Atomic mass }}\)
\(=\frac{5}{40}=\frac{1}{8}\) = 0.125 mole.

Question 49.
One mole of H₂O corresponds to _____.
(a) 22.4 litre at 1 atm and 250°C
(b) 6.023 × 10²³ atoms of hydrogen and 6.023 × 10²³ atoms of oxygen
(c) 18 g
(d) 1 g.
Answer:
(c) 18 g
Hint: One mole = Molar mass = 2 + 16 = 18 g.

Question 50.
Which one of the following has the maximum number of atoms?
(a) 18 g of H₂O
(b) 18 g of O₂
(c) 18 g of CO₂
(d) 18 g of CH₄.
Answer:
(d) 18 g of CH₄.
Hint:

Question 51.
The atomicity of K₂Cr₂O₇ is _____.
(a) 9
(b) 11
(c) 10
(d) 12.
Answer:
(b) 11

Question 52.
All noble gases are _____ molecules.
(a) diatomic
(b) triatomic
(c) mono atomic
(d) poly atomic.
Answer:
(c) mono atomic

Question 53.
The total number of atoms represented by the compound CuSO₄ . 5H₂O is ____.
(a) 27
(b) 21
(c) 5
(d) 8.
Answer:
(b) 21

Question 54.
Which one of the following represents the mass of 0.5 moles of water molecules?
(a) 18 g
(b) 1.8 g
(c) 9 g
(d) 4.5 g.
Answer:
(c) 9 g
\(\text { Mole }=\frac{\text { Mass }}{\text { Molecular mass }}\)
Mass = Mole × Molecular mass = 0.5 × 18 = 9 g.

Question 55.
The atomic mass of Calcium is 40. Calculate the number of moles in 16 g of Calcium.
(a) 0.4 mole
(b) 4 moles
(c) 640 moles
(d) \(\frac { 1 }{ 4 }\) mole.
Answer:
(a) 0.4 mole
Hint:
\(\text { Mole }=\frac{\text { Mass }}{\text { Atomic mass }}=\frac{16}{40}=\frac{8{}}{20}\) \(=\frac{4}{10}=0.4 \mathrm{mole}\).

Question 56.
If the atomic mass of sodium is 23 amu, then the mass of 3.011 × 10²³ sodium atoms is _____.
(a) 11.5 kg
(b) 23 g
(c) 0.5 mole
(d) 11.5 g.
Answer:
(d) 11.5 g.
Hint: Mass of 6.023 × 10²³ sodium atoms = 23 amu = 23 g.
∴ Mass of 3.011 × 10²³ sodium atoms
\(=\frac{23}{6.023 \times 10^{23}} \times 3.011 \times 10^{23}=11.5 \mathrm{g}\).

Question 57.
Which of the following will have maximum mass?
(а) 0.1 mole of NH₂
(b) 1022 atoms of carbon
(c) 1022 molecules of CO₂
(d) 1 g of Fe
Answer:
(a) 0.1 mole of NH₃
Hint:
(a) 0.1 mole of NH₃ has 6.023 × 10²³ atoms.
Mass of 1 mole of NH₃ = 17 g
Mass of 0.1 mole of NH₃ = 1.7 g.

(b) Mass of 1022 atoms of carbon
6.023 × 1023 c atoms mass = 12 g
1022 atoms of C has the mass
\(=\frac{12}{6.023 \times 10^{23}} \times 1022=2.036 \times 10^{-20} \mathrm{g}\).

(c) Mass of 1022 molecules of CO₂
CO₂ = molar mass = 44 g
6.023 × 10²³ CO₂ molecules has the mass = 44 g
∴ 1022 CO₂ molecules has the mass 44
\(=\frac{44}{6.023 \times 10^{23}} \times 1022=7.466 \times 10^{-20} \mathrm{g}\).

(d) 1 g of Fe
∴ (a) 1.7 g of NH₃ has the highest mass.

Question 58.
Which of the following correctly represents 360 g of water?
(i) 2 moles of water
(ii) 20 moles of water
(iii) 6.023 × 10²³ molecules of water
(iv) 1.2044 × 10²⁵ molecules of water
(a) (i) only
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv).
Answer:
(d) (ii) and (iv).
Hint: (i) 2 moles of water
Mass of 1 mole of water = 18 g
Mass of 2 moles of water = 18 × 2 = 36 g.

(ii) 20 moles of water
Mass of 1 mole of water = 18 g
Mass of 20 moles of water = 18 × 20 = 360 g.

(iii) 6.023 × 10²³ molecules of water = 1 mole = 18 g.

(iv) 1.2044 × 10²⁵ molecules of water
6.23 × 10²³ molecules of water = 1 mole
∴ 1.2044 × 10²⁵ molecules
\(=\frac{1}{6.023 \times 10^{23}} \times 1.2044 \times 10^{25}\)
= 20 moles.
∴ Mass of 20 moles = 20 × 18 = 360 g.
So (d) is correct.

Question 59.
Which of the following contains maximum number of molecules?
(a) 1 g of CO₂
(b) 1 g of N₂
(c) 1 g of H₂
(d) 1 g of CH₄
Answer:
(b) 1 g of H₂
Hint:
(a) 1 g of CO₂
No. of moles = \(\frac{\text { Mass }}{\text { Atomic mass }}\) (or) \(\frac{\text { Mass }}{\text { Molecular mass }}\)
No. of moles of 1 g of CO₂ = \(\frac{1}{44}\)
No. of molecules = \(\frac{1}{44}\) × 6.023 × 10²³
= 1.368 × 10²² molecules of CO₂.

(b) 1 g of N₂
No. of molecules = \(\frac{1}{28}\) × 6.023 × 10²³
= 2.151 × 10²² molecules of N₂.

(c) 1 g of H₂
No. of molecules = \(\frac{6.023 \times 10^{23} \times 1}{2}\)
= 3.011 × 10²³ molecules of H₂

(d) 1 g of CH₄
No. of molecules = \(\frac{6.023 \times 10^{23} \times 1}{16}\)
= 3.764 × 10²² molecules of CH₄
So (c) is the correct answer.

Question 60.
Which of the following pair is an example of isotopes?
\(\begin{array}{l}{\text { (a) } 21 \mathrm{Sc}^{45} \text { and }_{23} \mathrm{V}^{50}} \\ {\text { (b) }_{22} \mathrm{Ti}^{48} \text { and }_{22} \mathrm{Ti}^{50}} \\ {\text { (c) }_{22} \mathrm{Ti}^{50} \text { and }_{23} \mathrm{V}^{50}} \\ {\text { (d) }_{21} \mathrm{Sc}^{45} \text { and }_{22} \mathrm{Ti}^{50}}\end{array}\)
Answer:
(b) \(\text { (b) }_{22} \mathrm{Ti}^{48} \text { and }_{22} \mathrm{Ti}^{50}\).

II. Fill in the blanks.

Question 1.
Amedeo Avogadro put forward a hypothesis based on the relation between the number of _____ and the _____ of gases.
Answer:
Molecules, volume.

Question 2.
The molar volume of a gas at STP is _____ and the value of Avogadro Number is _____.
Answer:.
22.4.litres, 6.023 x 10²³.

Question 3.
Nitrogen and oxygen are _____ molecules whereas Helium and Neon are ____ molecules.
Answer:
Diatomic, monoatomic

Question 4.

1. _____ are the building blocks of matter.
2. ______ is a triatomic molecule.

Answer:

1. Atoms and molecules
2. Ozone

Question 5.
NH₃, H₂O are _____ molecules whereas N₂, O₂ are _____ molecules.
Answer:
Heteroatomic, Homoatomic

Question 6.
____ and ____ are polyatomic molecules.
Answer:
Phosphorous (P₄), Sulphur (S₈)

Question 7.

1. Atoms of the same element with same atomic number but a different mass number are called _____.
2. Atoms of different elements with the same number of neutrons are called _____.

Answer:

1. Isotopes
2. Isotones

Question 8.
Atomicity of Nitrogen is _____ whereas the atomicity of Helium is _____.
Answer:
2, 1.

Question 9.
Atoms of the same element with same atomic number but having different mass number are called _____.
Answer:
Isotopes.

Question 10.
Atoms of different elements with the same atomic mass but a different atomic number are called _____.
Answer:
Isobars.

Question 11.
Atoms of different elements having the same number of neutrons but a different atomic number and different mass number are called _____.
Answer:
Isotones.

Question 12.
_____ is the smallest particle that takes part in the chemical reaction.
Answer:
Atom.

Question 13.
Anything that has mass and occupies space is called _____.
Answer:
Matter.

Question 14.
Protons and neutrons have considerable mass, but _____ don’t have considerable mass.
Answer:
Atoms.

Question 15.
_____ is one-twelfth of the mass of C – 12 atom, an isotope of carbon which contains _____ protons and ____ neutrons.
Answer:
The atomic mass unit, 6, 6.

Question 16.
_____ are the building blocks of matter.
Answer:
Atoms.

Question 17.
The stable isotope of _____ is used as the standard for measuring the relative atomic mass of an element.
Answer:
Carbon C – 12.

Question 18.
Modem methods of determination of atomic mass by _____ use C – 12 as standard.
Answer:
Mass Spectrometry.

Question 19.
The relative atomic mass of sulphur is _____.
Answer:
32.

Question 20.
The average atomic mass of carbon is ______.
Answer:
12.011 amu.

Question 21.
The average atomic mass of an element becomes fractional due to the presence of ______.
Answer:
Isotopes.

Question 22.
_____ is the most abundant element in both the Earth’s crust and the human body.
Answer:
Oxygen.

Question 23.
Except for _____ atoms of most of the elements are found in the combined form with itself or atoms of other elements.
Answer:
Noble gases.

Question 24.
A molecule is a combination of two or more atoms held together by _____.
Answer:
Chemical bonds.

Question 25.
If the molecule is made of similar kind of atoms, it is called ______.
Answer:
Homo atomic molecule.

Question 26.
The molecule that consists of atoms of different elements is called _____.
Answer:
Hetero atomic molecule.

Question 27.
The number of _____ present in the molecule is called its atomicity.
Answer:
atoms.

Question 28.
The atomicity of ozone is _____.
Answer:
3.

Question 29.
The atomicity of hydrogen chloride is _____.
Answer:
2.

Question 30.
Water is a _____ molecule.
Answer:
Hetero triatomic.

Question 31.
One mole of an element contains ______ atoms and it is equal to its gram atomic mass.
Answer:
6.023 × 10²³

Question 32.
One mole of any gas occupies ______ or _____ at S.T.P.
Answer:
22.4 litre, 22400 ml.

Question 33.
The _____ is useful to determine the empirical formula and molecular formula.
Answer:
Percentage composition.

Question 34.
The percentage composition of elements is useful to determine _____ and _____.
Answer:
Empirical formula, molecular formula.

Question 35.
Avogadro’s law is in agreement with ______.
Answer:
Dalton’s atomic theory.

Question 36.
_____ determines the relation between molecular mass and vapour density.
Answer:
Avogadro’s law.

Question 37.
Relative molecular mass is equal to _____.
Answer:
2 × Vapour density.

Question 38.
Atomicity of sulphur is _____.
Answer:
8.

Question 39.
The metals Cu, Ag, Au are _____ elements.
Answer:
Monoatomic.

Question 40.
The atomicity of H₂SO₄ is ______.
Answer:
7.

Question 41.
Atomicity of an element is equal to _____.
Answer:
\(\frac{\text { Molecular mass }}{\text { Atomic mass }}\)

III. Match the following.

Question 1.

i. Monoatomic molecule	(a) Ozone
ii. Diatomic molecule	(b) Phosphorous
iii. Triatomic molecule	(c) Helium
iv. Polyatomic molecule	(d) Oxygen

Answer:
i – c, ii – d, iii – a, iv – b.

Question 2.

i. 22.4 litres	(a) Avogadro Number
ii. 6.023 × 1023	(b) Molar volume
iii. 2 × vapour density	(c) 1 mole
iv. Mass / Atomic mass	(d) Molecular mass

Answer:
i – b, ii – a, iii – d, iv – c.

Question 3.

i. \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\)	(a) Isotones
ii. \(_{6} \mathrm{Cl}^{13},_{7} \mathrm{N}^{14}\)	(b) Isobars
iii. \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\)	(c) E = mc2
iv. Einstein’s equation	(d) Isotopes

Answer:
i – d, ii – a, iii – b, iv – c.

Question 4.

i. H2O	(a) 180 g
ii. NH3	(b) 44 g
iii. CO2	(c) 17 g
iv. C6H12O6	(d) 18 g

Answer:
i – d, ii – c, iii – b, iv – a.

Question 5.

i. NH3, CH4	(a) Polyatomic molecule
ii. O2, N2	(b) Monoatomic molecule
iii. He, Ne	(c) Heteroatomic molecule
iv. Sulphur	(d) Diatomic molecule

Answer:
i – c, ii – d, iii – b, iv – a.

Question 6.

i. F2	(a) Polyatomic molecule
ii. O3	(b) Monoatomic molecule
iii. P4	(c) Diatomic molecule
iv. He	(d) Triatomic molecule

Answer:
i – c, ii – d, iii – a, iv – b.

Question 7.

i. H2	(a) Hetero diatomic molecule
ii. HCl	(b) Monoatomic molecule
iii. H2O	(c) Homo diatomic molecule
iv. Ne	(d) Hetero triatomic molecule

Answer:
i – c, ii – a, iii – d, iv – b.

Question 8.

i. Isotopes	(a) S8, P4
ii. Isobars	(b) \(_{6} \mathrm{C}^{13},_{7} \mathrm{N}^{14}\)
iii. Isotones	(c) \(_{1} \mathrm{H}^{1},_{1} \mathrm{H}^{2},_{1} \mathrm{H}^{3}\)
iv. Polyatomic molecule	(d) \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\)

Answer:
i – c, ii – d, iii – b, iv – a.

Question 9.

i. H2O	(a) 16
ii. CO2	(b) 18
iii. C6H12O6	(c) 44
iv. CH4	(d) 180

Answer:
i – b, ii – c, iii – d, iv – a

Question 10.

i. 22 g of CO2	(a) 2 moles
ii. 18 g of H2O	(b) 4 moles
iii. 360 g of Glucose	(c) 0.5 mole
iv. 64 g of CH4	(d) 1 mole

Answer:
i – c, ii – d, iii – a, iv – b.

IV. State whether true or false. If false, give the correct statement.

Question 1.
Isotopes are the atoms of the same element may not be similar in all respects.
Answer:
True.

Question 2.
Isobars are the atoms of the different elements with the same atomic number and different mass numbers.
Answer:
False.
Correct statement: Isobars are the atoms of the different elements with the same mass number but a different atomic number.

Question 3.
Isotones are the atoms of different elements with the same number of neutrons.
Answer:
True.

Question 4.
The number of molecules present in one mole of an element is called atomicity of an element.
Answer:
False.
Correct statement: The number of atoms present in one molecule of an element is called the atomicity of an element.

Question 5.
Avogadro’s hypothesis is used in the deduction of atomicity of elementary gases.
Answer:
True.

Question 6.
The volume of a gas at a given temperature and pressure is proportional to the number of particles.
Answer:
True.

Question 7.
The value of Gram molar volume at STP is 11.2 litres.
Answer:
False.
Correct statement: The value of Gram molar volume at STP is 22.4 litres.

Question 8.
The atomicity of nitrogen, oxygen and hydrogen is two.
Answer:
True.

Question 9.
Atoms and molecules are the building blocks of matter.
Answer:
True.

Question 10.
The atoms of certain elements such as hydrogen, oxygen and nitrogen have an independent existence.
Answer:
False.
Correct statement: The atoms of certain elements such as hydrogen, oxygen and nitrogen do not have an independent existence.

Question 11.
A molecule is the simplest structural unit of an element or compound which contains one or more atoms.
Answer:
True.

Question 12.
Phosphorous and sulphur are monoatomic molecules.
Answer:
False.
Correct statement: Phosphorous and sulphur are polyatomic molecules.

Question 13.
H₂O, NH₃, CH₄ are examples of homoatomic molecules.
Answer:
False.
Correct statement: H₂O, NH₃, CH₄ are examples of heteroatomic molecules.

Question 14.
An atom of one element can be transmuted into an atom of other element is known as artificial transmutation.
Answer:
True.

Question 15.
The molecule is the smallest particle that takes part in a chemical reaction.
Answer:
False.
Correct statement: Atom is the smallest particle that takes part in a chemical reaction.

Question 16.
The sum of the number of protons and neutrons of an atom is called Atomic number.
Answer:
False.
Correct statement: The sum of the number of protons and neutrons of an atom is called mass number.

Question 17.
The stable isotope of carbon (C – 12) with atomic mass 12 is used as the standard for measuring the relative atomic mass of an element.
Answer:
True.

Question 18.
The gram atomic mass of oxygen is 16 g.
Answer:
True.

Question 19.
Silicon is the most abundant element in the Earth’s crust.
Answer:
False.
Correct statement: Oxygen is the most abundant element in the Earth’s crust.

Question 20.
Except for noble gases, atoms of most of the elements are found in the combined form.
Answer:
True.

Question 21.
The number of atoms present in the molecule is called the Avogadro number.
Answer:
False.
Correct statement: The number of atoms present in the molecule is called its Atomicity.

Question 22.
O₂, N₂, H₂, Cl₂, Br₂, F₂, I₂ are hetero diatomic molecules.
Answer:
False.
Correct statement: O₂, N₂, H₂, Cl₂, Br₂, F₂, I₂ are homo diatomic molecules.

Question 23.
Water is an example of Hetero triatomic molecule.
Answer:
True.

Question 24.
One molecule of an element contains 6.023 × 10²³ atoms and it is equal to its gram atomic mass.
Answer:
True.

Question 25.
An equal volume of all gases under similar conditions of temperature and pressure contain a different number of molecules.
Answer:
False.
Correct statement: Equal volume of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Question 26.
The mathematical representation of Avogadro’slawisV/n=Constant(or)Vccn(or) V = Constant × n.
Answer:
True.

Question 27.
The molecular formula of gases can be derived using Avogadro’s law.
Answer:
True.

Question 28.
The number of moles of a sample that contains 12.046 x 10²³ atoms of iron is 2.
Answer:
True.

Question 29.
The volume occupied by 14 g of Nitrogen gas is 22.4 litres.
Answer:
False.
Correct statement: The volume occupied by 14 g of Nitrogen gas is 11.2 litres.

Question 30.
Avogadro’s law determines the relation between molecular mass and absolute density.
Answer:
False.
Correct statement: Avogadro’s law determines the relation between molecular mass and vapour density.

V. Assertion and Reason

Question 1.
Assertion (A): C₁₂H₂₂O₁₁ is not a simple ratio.
Reason (R): The ratio of atoms in a molecule may be fixed and integral but may not be simple.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 2.
Assertion (A): \(_{6} \mathrm{C}^{13}\) and \(_{7} \mathrm{N}^{4}\) are called Isotones.
Reason (R): Isotones are the atoms of the different elements with different atomic number but the same mass number.
(a) Both (A) and (R) are correct
(b)Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 3.
Assertion (A): Nitrogen, oxygen and hydrogen are diatomic molecules.
Reason (R): N₂, O₂, H₂ contain two atoms in one molecule and so they are a diatomic molecule.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 4.
Assertion (A): Atoms and molecules are the building blocks of matter.
Reason (R): Atom is the ultimate particle of an element which may or may not have an independent existence.
(a) Both (A) and (R) are wrong]
(b) (A) is correct but (R) does not explain (A)
(c) Both (A) and (R) are correct
(d) (A) is wrong but (R) is correct.
Answer:
(c) Both (A) and (R) are correct

Question 5.
Assertion (A): Hydrogen, Oxygen and Ozone are called homoatomic molecules.
Reason (R): Homoatomic molecules are made up of atoms of the same element.
(a) Both (A) and (R) are wrong
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are correct.
Answer:
(d) Both (A) and (R) are correct.

Question 6.
Assertion (A): Water, Ammonia (H₂O, NH₃) are heteroatomic molecules.
Reason (R): Most of the elementary gases and compounds consist of atoms of the same element.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 7.
Assertion (A): 18 g water contains Avogadro number (6.023 × 10²³) of particles.
Reason (R): 18 g of water is the molecular mass (or) 1 mole of water. One mole is defined as the amount of the substance which contains 6.023 × 10²³ number of particles.
(a) (A) is correct and (R) explains (A)
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong
Answer:
(a) (A) is correct and (R) explains (A)

Question 8.
Assertion (A): Atoms of the same element may not be similar in all respects.
Reason (R): Atoms of the same element have the same atomic number but a different number of neutrons.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) Both (A) and (R) are wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) (A) is correct but (R)is wrong

Question 9.
Assertion (A): The atomicity of ozone is three.
Reason (R): 1 molecule of ozone contains 3 atoms of oxygen.
(a) Both (A) and (R) are correct
(b) Both (A) and(R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): \(_{1} \mathrm{H}^{1}, \quad_{1} \mathrm{H}^{2},_{1} \mathrm{H}^{3}\) are the isotopes of hydrogen.
Reason (R): The atoms of the same element with the same mass number but different at numbers are called isotopes.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Assertion (A) & Reason (R):
(i) (A) and (R) are correct. (R) explain (A)
(ii) (A) is correct (R) is wrong
(iii) (A) is wrong (R) is correct
(iv) (A) and (R) are correct. (R) does not explain (A).

Question 11.
Assertion (A): An atom is no longer indivisible.
Reason (R): The subatomic particles protons, electrons and neutrons were discovered.
Answer:
(i) (A) and (R) are correct. (R) explain (A)

Question 12.
Assertion (A): \(_{18} \mathrm{Ar}^{40}\) and \(_{20} \mathrm{Ca}^{40}\) are isobars.
Reason (R): They have the same atomic mass but a different atomic number.
Answer:
(i) A) and (R) are correct; (R) explain (A)

Question 13.
Assertion (A): \(_{17} \mathrm{Cl}^{35}\) and \(_{17} \mathrm{Cl}^{37}\) are isotones.
Reason (R): Atoms of the same element have the same atomic number but a different mass number.
Answer:
(iii) (A) is wrong (R) is correct

Question 14.
Assertion (A): NH₃, H₂O, HCl are heteroatomic molecules.
Reason (R): The molecule that consists of atoms of different elements is called heteroatomic molecules.
Answer:
(i) (A) and (R) are correct. (R) explain (A)

Question 15.
Assertion (A): \(_{6} \mathrm{C}^{13}\) and \(_{7} \mathrm{N}^{14}\) are called isotones.
Reason (R): Atoms of different elements having the same number of neutrons, but a different atomic number and different mass number are called isotones.
Answer:
(i) (A) and (R) are correct. (R) explain (A)

VI. Short Answer Questions.

Question 1.
What are isotopes? Give an example.
Answer:
Atoms of the same element that have same atomic number but different mass number are called isotopes.
e.g., \(_{1} \mathrm{H}^{1},_{1} \mathrm{H}^{2},_{1} \mathrm{H}^{3}\).

Question 2.
State Avogadro Hypothesis.
Answer:
Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Question 3.
What are isotones? Give an example.
Answer:
Atoms of different elements having the same number of neutrons but a different atomic number and different mass numbers are called isotones.
e.g., \(_{6} \mathrm{C}^{13},_{7} \mathrm{N}^{14}\).

Question 4.
Define Mole.
Answer:
Mole is defined as the amount of substance that contains as many specified elementary particles as the number of atoms in 12 g of C-12 isotope.
It is also defined as the amount of substance which contains Avogadro number (6.023 × 10²³) of particles.

Question 5.
Define

1. Atomic number
2. Mass number

Answer:

1. The atomic number of an element is the number of protons or number of neutrons and electrons present in it.
2. The mass number is the sum of the number of protons and neutrons in an atom.

Question 6.
How many grams are there in
(i) 5 moles of H₂O
Answer:
5 moles of H₂O = 5 × 18 = 90 g

(ii) 1 mole of Glucose (C₆H₁₂O₆)
Answer:
1 mole of Glucose (C₆H₁₂O₆) = 180 g

Question 7.
Define molecule.
Answer:
A molecule is a combination of two or more atoms held together by the strong chemical force of attraction, i.e. Chemical bonds.

Question 8.
What is homo atomic molecule? Give two examples.
Answer:
If the molecule is made of similar kind of atoms, then it is called homoatomic molecule. e.g. H₂, Cl₂

Question 9.
What is a heteroatomic molecule? Give two examples.
Answer:
The molecule that consists of atoms of different elements is called a heteroatomic molecule. e.g. HCl, H₂O

Question 10.
Consider the following and classify them on the basis of their atomicity.
H₂, CCl₄, O₃, BF₃, HCl, HNO₃, C₁₂H₂₂O₁₁, NO, Cl₂, He, Au, P₄

• Monoatomic molecule – He, Au
• Homo diatomic molecule – H₂, Cl₂
• Homo triatomic molecule – O₃
• Homo polyatomic molecule – P₄
• Hetero diatomic molecule – HCl, NO
• Hetero polyatomic molecule – CCl₄, BF₃, HNO₃, C₁₂H₂₂O₁₁.

Question 11.
Define Relative molecular mass.
Answer:
The Relative molecular mass of a molecule is the ratio between the mass of one molecule of the substance to 1 / 12th mass of an atom of Carbon – 12 isotope.

Question 12.
Define Mole.
Answer:
The mole is the amount of the substance that contains as many elementary entities as there are atoms in exactly 12 g of the carbon – 12 isotope.

Question 13.
Define the Avogadro number.
Answer:
The actual number of atoms in 12 g of carbon – 12 is called the Avogadro number.
It is equal to 6.023 × 10²³ (N_A).

Question 14.
What is meant by percentage composition? What is its use?
Answer:
The percentage composition of a compound represents the mass of each element present in 100 g of the compound. It is useful to determine the empirical formula and molecular formula.

Question 15.
State Avogadro hypothesis (or) Avogadro’s Law.
Answer:
The Avogadro’s law states that “equal volume of all gases under similar conditions of temperature and pressure contain the equal number of molecules”.
[V ∝ n].

Question 16.
What are the applications of Avogadro’s Law?
Answer:

• It explains Gay – Lussac’s law.
• It helps in the determination of atomicity of gases.
• The molecular formula of gases can be derived using Avogadro’s law.
• It determines the relation between molecular mass and vapour density.
• It helps to determine the gram molar volume of all gases, (i.e, 22.4 litres at S.T.P).

Question 17.
How is Average atomic mass calculated?
Answer:
The average atomic mass of an element is calculated by adding the masses of its isotopes, each multiplied by their natural abundance on the Earth.

Question 18.
Define Vapour density.
Answer:
The vapour density is defined as the ratio between the masses of equal volumes of a gas (or vapour) and hydrogen under the same condition.

Question 19.
Write the relationship between

1. Atomicity and Molecular mass
2. Molecular mass and Vapour density.

Answer:

1. Atomicity = \(\frac{\text { Molecular mass }}{\text { Atomic mass }}\)
2. Molecular mass = 2 × Vapour density

Question 20.
Distinguish between isotopes and isobars.
Answer:

Isotopes	Isobars
The atoms of the same element with same atomic number (Z) but different mass number (A) are called isotopes. e.g. \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\)	The atoms of the different element with the same mass number (A) but different atomic number (Z) are called isobars. e.g. \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\)

Question 21.
What are the types of molecules? Give an example for each type?
Answer:
Molecules are of two types:

• Homoatomic molecule: The molecules which are made up of atoms of the same element are called Homoatomic molecule, e.g., N₂, O₂, H₂
• Heteroatomic molecule: The molecules which are made up of atoms of different elements are called Heteroatomic molecule, e.g., NH₃, H₂O, CH₄

VII. HOT Questions.

Question 1.
Calculate the mass of CO₂ which contains the same number of molecules as are contained in 40 g of SO₂.
Answer:
Gram molecular mass of SO₂ = 32 + 2(16)
= 64 g
No. of moles of SO₂ = \(\frac{GivenMass}{Mol.Mass}\)
= \(\frac{40}{64}\) = 0.625 moles
∵ Equal moles contains equal number of molecules.
Mass of CO₂ which contains the same number of molecules,
= 0.625 × mol. mass of CO₂
= 0.625 × 44
= 27.5 g

Question 2.
A flask P contains 0.5 moles of oxygen gas. Another flask Q contains 0.4 moles of ozone gas. Which of the two flasks contains greater number of oxygen atoms?
Answer:
1 molecule of oxygen (O₂) = 2 atoms of oxygen
1 molecule of ozone (O₃) = 3 atoms of oxygen
In flask P:
1 mole of oxygen gas = 6.022 × 10²³ molecules
0.5 mole of oxygen gas = 6.022 × 10²³ × 0.5 molecules
= 6.022 × 10²³ × 0.5 × 2 atoms
= 6.022 × 10²³ atoms.

In flask Q:
1 mole of ozone gas = 6.022 × 10²³ molecules
0.4 mole of ozone gas = 6.022 × 10²³ × 0.4 molecules
= 6.022 × 10²³ × 0.4 × 3 atoms
= 7.23 × 10²² atoms
Flask Q has a greater number of oxygen atoms as compared to the flask P.

Question 3.
Chlorophyll, the green pigment of plants responsible for photosynthesis contain 2.68% of Mg by weight. Calculate the number of magnesium atoms in 20 g of chlorophyll.
Answer:
The weight % of Mg as 2.68
i.e.,100 g of chlorophyll contains 2.68 g of Mg
∴ 2 g of chlorophyll will contain Mg
\(\frac{2.68}{100}\) × 20
= 0.5369
1 mole of Mg = 24 g = 6.023 × 10²³ atoms
∴ 0.0536 g of Mg will have = \(\frac{6.023×10^{23}}{24}\) × 0.536
= 0.1345 × 10²³ atoms of Mg = 1.345 × 10²²
Number of Magnesium atoms present in 20 g of chlorophyll is 1.345 × 10²²

Question 4.
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 C – atoms = 6 moles
2. 1 mole of C₂H₆ contains 6 moles of hydrogen atoms
   3 moles of C₂H₆ will contain H-atoms = 18 moles
3. 1 mole of C₂H₆ contains Avogadro’s number. i.e., 6.023 × 10²³ molecules.
   3 moles of C₂H₆ will contain ethane molecules = 3 × 6.023 × 10²³ = 18.06 × 10²³ molecules.

Question 5.
If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour could be produced?
Answer:
H₂ and O₂ react according to the equation
H_{2 (g)} + O_{2 (g)} → 2H₂O _(g)
Thus, 2 volumes of H₂ react with 1 volume of O₂ to produce 2 volumes of water vapour.
Hence, 10 volumes of H₂ will react completely with 5 volumes of O₂ to produce 10 volumes of water vapour.

VIII. Long Answer Questions.

Question 1.
What are the differences between atoms and molecules?
Answer:

Atom	Molecule
An atom is the smallest particle of an element	A molecule is the smallest particle of an element or compound.
Atom does not exist in the free state except in a noble gas	The molecule exists in the free state
Except some of the noble gas, other atoms are highly reactive	Molecules are less reactive
Atom does not have a chemical bond	Atoms in a molecule are held by chemical bonds
Example: Na	Example: N2

Question 2.
Write the applications of Avogadro’s Law.
Answer:
(i) It explains Gay-Lussac’s law.
(ii) It helps in the determination of atomicity of gases.
(iii) Molecular formula of the gases can be derived.
(iv) It determines the relation between molecular mass and vapour density.
(v) It helps to determine gram molar volume of all gases

Question 3.
State and explain the applications of Avogadro’s law.
(OR)
Give any two applications of Avogadro’s law.
(OR)
Write any three applications of Avogadro’s law.
Answer:
Avogadro’s law: Equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Applications of Avogadro’s law:

• It is used to determine the atomicity of gases.
• It is helpful in determining the molecular formula of gaseous compounds.
• It establishes the relationship between the vapour density and molecular mass of a gas.
• It gives the value of the molar volume of gases at STP. Molar volume of a gas at STP = 22.4 litres.
• It explains Gaylussac’s law effectively.

Question 4.
Explain the classification of molecules based on atomicity.
Answer:
In accordance with the number of atoms present in the molecules, they are classified as monoatomic, diatomic, triatomic and polyatomic molecules showing that they contain one, two, three or more than 3 atoms respectively.

Atomicity	Number of atoms per molecule	Example
Monoatomic molecule	1	Helium (He), Neon (Ne) metals (Fe, Cu)
Diatomic molecule	2	Hydrogen (H2), Chlorine (Cl2)
Triatomic molecule	3	Ozone (O3)
Polyatomic molecule	>3	Phosphorous (P4), Sulphur (S8)

Question 5.
A compound made up of two elements A and B has A = 70%, B = 30%. Their relative number of moles in the compound are 1.25 and 1.88. Calculate.
(a) Atomic masses of the elements A and B.
(b) The molecular formula of the compound, if its molecular mass is found to be 160.
Answer:

Elements	Relative no. of moles	Simplest molar ratio	Simplest whole no. molar ratio
A	1.25	\(\frac{1.25}{1.25}=1\)	2
B	1.88	\(\frac{1.88}{1.25}=1.5\)	3

(a) Atomic mass of A = \(\frac{70}{1.25}\) = 56
Atomic mass of B = \(\frac{30}{1.88}\) = 16

(b) The molecular mass of the compound = 160
The molecular formula of the compound = Fe₂O₃

IX. Solve the following problems.

Question 1.
Calculate the gram molar mass of the following.
(a) NaOH
(b) C₁₂H₂₂O₁₁
(c) H₃PO₄
(Atomic mass of Na – 23, O -16, H – 1, C – 12, P – 31)
Answer:
(a) NaOH (Sodium hydroxide)
GMM = 23 + 16 + 1
= 40 g
Gram molar mass of NaOH = 40 g

(b) C₁₂H₂₂O₁₁ (Sucrose)
GMM = 12 × 12 + 22 × 1 + 11(16)
= 342 g
Gram molar mass of sucrose = 342 g

(c) H₃ PO₄ (Phosphoric acid)
GMM = 3(1) + 1(31)+ 4(16) = 98 g
Gram molar mass of Phosphoric acid = 98 g.

Question 2.
Calculate the percentage composition of oxygen and hydrogen by taking the example of H₂O
Solution:
Mass % of an element = \(\frac{\text { Mass of that element in the compound }}{\text { Molar mass of the compound }} \times 100\)
Now, Molar mass of H₂O = 2(1) + 16 = 18 g
Mass % of Hydrogen = \(\frac{2}{18} \times 100\) = 11.11 %
Mass % of Oxygen = \(\frac{16}{18} \times 100\) = 88.89 %.

Question 3.
What is the mass of 1 atom of Gold? (At. mass of Au = 197)
Answer:
The mass of 6.023 × 10²³ atoms of Gold = 197 g
∴ The mass of 1 atom of gold = \(\frac{197}{6.023×10^{23}}\) × 1
= 3.27 × 10^-22 g

Question 4.
Find the gram molecular mass of the following from the data given:
(i) H₂O
(ii) CO₂
(iii) NaOH
(iv) NO₂
(v) H₂SO₄

Element	Symbol	Atomic No.	Atomic Mass
Hydrogen	H	1	1
Carbon	C	6	12
Oxygen	O	8	16
Nitrogen	N	7	14
Sodium	Na	11	23
Sulphur	S	16	32

Solution:
(i) H₂O
Atomic mass of 2(H) = 2 × 1 = 2
Atomic mass of 1(O) = 1 × 16 = 16
Molecular mass of H₂O = 2 + 16 = 18

(ii) CO₂
Atomic mass of 1(C) = 1 × 12 = 12
Atomic mass of 2(O) = 2 × 16 = 32
Molecular mass of CO₂ = 12 + 32 = 44 g

(iii) NaOH
Atomic mass of 1(Na) = 1 × 23 = 23
Atomic mass of 1(O) = 1 × 16 = 16
Atomic mass of 1(H) 1 × 1 = 1
Molecular mass of NaOH = 23 + 16 + 1 = 40 g

(iv) NO₂
Atomic mass of 1(N) = 1 × 14 = 14
Atomic mass of 2(O) = 2 × 16 = 32
Molecular mass of NO₂ = 14 + 32 = 46 g.

(v) H₂SO₂₄
Atomic mass of 2(H) = 2 × 1= 2
Atomic mass of 1(S) = 1 × 32 = 32
Atomic mass of 4(O) = 4 × 16 = 64
Molecular mass of H₂SO₄ = 64 + 32 + 2 = 98 g.

Question 5.
Complete the table given below.

Element	Atomic Mass	Molecular Mass	Atomicity
Chlorine	35.5	71	–
Ozone	–	45	3
Sulphur	32	–	8

Solution:

Element	Atomic Mass	Molecular Mass	Atomicity
Chlorine	35.5	71	2
Ozone	16	48	3
Sulphur	32	256	8

Question 6.
Fill in the blanks using the given data:
The formula of Calcium oxide is CaO. The atomic mass of Ca is 40, Oxygen is 16 and Carbon is 12.

• 1 mole of Ca (….. g) and 1 mole of the Oxygen atom (…… g) combine to form mole of CaO (….. g).
• 1 mole of Ca (…… g) and 1 mole of C (…… g) and 3 moles of the Oxygen atom (…… g) combine to form 1 mole of CaCO₃ (…… g).

Solution:

• 1 mole of Ca (40 g) and 1 mole of the Oxygen atom (16 g) combine to form 1 mole of CaO (56 g).
• 1 mole of Ca (40 g) and 1 mole of C (12 g) and 3 moles of the Oxygen atom (48 g) combine to form 1 mole of CaCO₃ (100 g).

Question 7.
Calculate the average atomic mass of naturally occurring magnesium using the following data.
Mg – 24 = 78.99% , Mg – 25 = 10%, Mg – 26 = 11.01%
Answer:
Average atomic mass of Magnesium = atomic mass of Mg – 24 × % + atomic mass of Mg – 25 × % + atomic mass of Mg – 26 × %
= 24 × \(\frac{78.99}{100}\) + 25 × \(\frac{10}{100}\) + 26 × \(\frac{11.01}{100}\)
= 18.9576 + 2.5 + 2.8626
= = 24.3202 amu
∴ Average atomic mass of Magnesium is 24.3202 amu

Question 8.
Analyse the table and fill in the blanks.

Gas	Atomic mass	Molecular mass	Atomicity
Ozone	16	48	–
Nitrogen	14	–	2

Solution:

Gas	Atomic mass	Molecular mass	Atomicity
Ozone	16	48	3
Nitrogen	14	28	2

Question 9.
Analyse the table and fill in the blanks.

Substance	Mass	No.of moles
(a) Al	81 g	–
(b) Fe	–	0.5

Solution:

Substance	Mass	No.of moles
(a) Al	81 g	3
(b) Fe	27.95 g	0.5

Question 10.
When ammonia reacts with hydrogen chloride gas, it produces white fumes of ammonium chloride. The volume occupied by NH₃ in glass bulb A is three times more than the volume occupied by HCl in glass bulb B at STP.

(i) How many moles of ammonia are present in glass bulb A?
(ii) How many grams of NH₄Cl will be formed when the stopper is opened? (Atomic mass of N = 14, H = 1, Cl = 35.5)
(iii) Which gas will remain after completion of the reaction?
(iv) Write the chemical reaction involved in this process.
Solution:
(i) Capacity of NH₃ bulb = 67.2 litre
22.4 litre of NH₃ = 1 mole
67.2 litre of NH₃ = \(\frac{1}{22.4} \times 67.2\) = 3 moles of NH₃

(ii) Atomic mass of 1(N) = 1 × 14 = 14 g
Atomic mass of 4(H) = 4 × 1 = 4 g
Atomic mass of 1(Cl) = 1 × 35.5 = 35.5 g
Mass of NH₄Cl = 53.5 g.

(iii) NH3 (Ammonia) gas will remain after the completion of the reaction.

(iv) Chemical equation of the reaction
NH₃ (Ammonia) + HCl (Hydrochloric acid) → NH₄Cl (Ammonium chloride)

Question 11.
Nitroglycerine is used as an explosive. The equation for the explosive reaction is
4C₃H₅((NO₃))₃ (l) → 12CO₂ (g) + 10H₂O (l) + 6N₂ (g) + O₂ (g)
(Atomic mass of C = 12, H = 1, N = 14, O = 16)
(i) How many moles does the equation show for:
(a) Nitroglycerine
(b) gas molecules produced?
(ii) How many moles of gas molecules are obtained from 1 mole of nitroglycerine?
(iii) What is the mass of 1 mole of nitroglycerine?
Solution:
(i) 4 moles of Nitroglycerine

(ii) 4 moles of Nitroglycerine produce 19 moles of gas molecules
1 mole of Nitroglycerine produces 19 / 4 = 4.75 moles

(iii) Mass of 1 mole of Nitroglycerine C₃H₅(NO₃)₃
Atomic mass of C = 12
Atomic mass of 3(C) = 3 × 12 = 36
Atomic mass of 5(H) = 5 × 1 = 5
Atomic mass of 3(N) = 3 × 14 = 42
Atomic mass of 9(O) = 9 × 16 = 144
Mass of 1 mole of Nitroglycerine = 227 g

Question 12.
Sodium bicarbonate breaks down on heating:
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
(Atomic mass of Na = 23, C = 12, H = 1, O = 16)
(i) How many moles of sodium bicarbonate are there in this equation?
(ii) What is the mass of sodium bicarbonate used in this equation?
(iii) How many moles of carbon dioxide are there in this equation?
Solution:
\(2 \mathrm{NaHCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \uparrow\)
(i) 2 moles of NaHCO₃ (sodium bicarbonate) are there in the above equation.

(ii) Mass of sodium bicarbonate in this equation is mass of 2 moles of NaHCO₃.
Atomic mass of 1(Na) = 1 × 23 = 23 g
Atomic mass of 1(H) = 1 × 1 = 1 g
Atomic mass of 1(C) = 1 × 12 = 12 g
Atomic mass of 3(O) = 3 × 16 = 48 g
Mass of 1 mole of NaHCO₃ = 84 g
Mass of 2 moles of NaHCO₃ = 84 × 2 = 168 g.

(iii) Number of moles of CO₂ in this equation = 1 mole.

Question 13.
40 g of calcium was extracted from 56 g of calcium oxide (Atomic mass of Ca = 40, O = 16)
(i) What mass of oxygen is there in 56 g of calcium oxide?
(ii) How many moles of oxygen atoms are there in this?
(iii) How many moles of calcium atoms are there in 40 g of calcium?
(iv) What mass of calcium will be obtained from 1000 g of calcium oxide?
Solution:
(i) Mass of CaO = 56 g
Mass of Ca = 40 g
Mass of oxygen = 56 – 40 = 16 g

(ii) \(\frac{\text { No of moles of oxygen atom }}{\text { Mole }=\text { mass/atomic mass }}=\frac{16}{16}=1 \text { mole }\)
1 mole of oxygen atom.

(iii) \(\frac{\text { No of moles of calcium }}{\text { Mole }=\text { mass/atomic mass }}=\frac{40}{40}=1 \text { mole }\)
1 mole of calcium atom.

(iv) 56 g calcium oxide gives 40 g of calcium
1000 g of calcium oxide give = \(\frac{40}{56} \times 1000\)
= 714.285 g of calcium
= 714.29 g of calcium.

Question 14.
How many grams are there in the following?
(i) 1 mole of chlorine molecule, Cl₂
(ii) 2 moles of sulphur molecules, S₈
(iii) 4 moles of ozone molecules, O₃
(iv) 2 moles of nitrogen molecules, N₂
Solution:
(i) 1 mole of chlorine molecule Cl₂
Atomic mass of chlorine = 35.5 g
Mass of 1 mole of chlorine = Atomic mass × Atomicity = 35.5 × 2 = 71 g.

(ii) 2 moles of sulphur molecules S₈
Atomic mass of S₈ = 8 × 32 = 256 g
Mass of 2 moles of S₈ = Atomic mass × Number of moles = 256 × 2 = 512 g.

(iii) 4 moles of ozone molecule O₃
Atomic mass of O₃ = 3 × 16 = 48 g
Mass of 4 moles of ozone = 48 × 4 = 192 g.

(iv) 2 moles of Nitrogen molecule N₂
Atomic mass of N₂ = 2 × 14 = 28 g
Mass of 2 moles of Nitrogen = 28 × 2 = 56 g.

Question 15.
Find how many moles of atoms are there in:
(i) 2 g of nitrogen
(ii) 23 g of sodium
(iii) 40 g of calcium
(iv) 1.4 g of lithium
(v) 32 g of sulphur.
Solution:
(i) 2 g of nitrogen
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}\)
Atomic mass of Nitrogen =14; Mass of Nitrogen = 2 g
Number of moles = \(\frac{2}{14}\) = 0.142 moles of Nitrogen.

(ii) 23 g of sodium
Atomic mass of sodium = 23
Mass of sodium = 23 g
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{23}{23}\) = 1 mole of sodium.

(iii) 40 g of calcium
Atomic mass of calcium = 40
Mass of calcium = 40 g
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{40}{40}\) = 1 mole of calcium.

(iv) 1.4 g of lithium
Atomic mass of lithium = 7
Mass of lithium = 1.4 g
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{1.4}{7}\) = 0.2 mole of lithium.

(v) 32g of sulphur
Atomic mass of sulphur = 32
Mass of sulphur = 32 g
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{32}{32}\) = 1 mole of sulphur.

Question 16.
Find the atomicity of chlorine, if its atomic mass is 35.5 and its molecular mass is 71.
Solution:
Atomicity = \(\frac{\text { Molecular mass }}{\text { Atomic mass }}\)
Atomicity of chlorine = \(\frac{71}{35.5}\) = 2.

Question 17.
Find the atomicity of ozone if its atomic mass is 16 and its molecular mass is 48.
Solution:
Atomicity = \(\frac{\text { Molecular mass }}{\text { Atomic mass }}\)
Atomicity of chlorine = \(\frac{48}{16}\) = 3.

Question 18.
How many atoms are present in 5 moles of oxygen?
Solution:
One mole of oxygen contains 6.023 × 1023 atoms
5 moles of oxygen contain = 5 × 6.023 × 10²³
= 30.115 × 10²³
= 3.0115 × 10²⁴ atoms.

Question 19.
Calculate the number of moles in

1. 81 g of Aluminium
2. 4.6 g of sodium
3. 5.1 g of ammonia
4. 90 g of water
5. 2 g of NaOH.

Solution:
No of moles = \(\frac{\text { Given mass }}{\text { Atomic mass }}\)

1. No. of moles of Aluminium = \(\frac { 81 }{ 27 }\) = 3 moles of aluminium
2. No. of moles of Sodium = \(\frac { 4.6 }{ 23 }\) = 0.2 moles of sodium
3. No. of moles of Ammonia = \(\frac { 5.1 }{ 17 }\) = 0.3 moles of ammonia
4. No. of moles of Water = \(\frac { 90 }{ 18 }\) = 5 moles of water
5. No. of moles of NaOH = \(\frac { 2 }{ 40 }\) = 0.05 moles of NaOH

Question 20.
Calculate the mass of 0.5 moles of iron.
Solution:
Mass = Atomic mass × number of moles
Mass of iron = 55.9 × 0.5 = 27.95 g.

Question 21.
Find the mass of 2.5 moles of oxygen atoms.
Solution:
Mass = Atomic mass × number of moles
Mass of oxygen = 16 × 2.5 = 40 g.

Question 22.
Calculate the number of molecules in 11 g of CO₂.
Solution:
Gram molecular mass of CO₂ = 44 g
Number of molecules = \(\frac{\text { Avogadro number } \times \text { given mass }}{\text { Gram molecular mass }}\)
= \(\frac{6.023 \times 10^{23} \times 11}{44}\)
= 1.51 × 10²³ CO₂ molecules.

Question 23.
Calculate the number of molecules in 360 g of glucose.
Solution:
Number of molecules = \(\frac{\text { Avogadro number } \times \text { given mass }}{\text { Gram molar mass }}\)
Gram molar mass of glucose (C₆H₁₂O₆) = (6 × 12) + (12 × 1) + (6 × 16)
= 72 + 12 + 96
= 180 g
Number of molecules = \(\frac{6.023 \times 10^{23} \times 360}{180}\)
= 6.023 × 10²³ × 2
= 12.046 × 10²³ molecules.
(or)
1.2046 × 10²⁴ glucose molecules.

Question 24.
Calculate the mass of 18.069 × 10²³ molecules of SO₂?
Solution:
Mass of the substance = \(\frac{\text { Gram molecular mass } \times \text { Number of particles }}{\text { Avogadro number }}\)
Gram molecular mass of SO₂ = 32 + 16(2) = 64 g
Mass of SO₂ = \(\frac{64 \times 18.069 \times 10^{23}}{6.023 \times 10^{23}}\)
= 64 × 3 = 192 g.

Question 25.
Calculate the mass of glucose in 2 × 10²⁴ molecules.
Solution:
Gram molecular mass of glucose = 180 g
Mass of glucose = \(\frac{180 \times 2 \times 10^{24}}{6.023 \times 10^{23}}\) = 597.7 g

Question 26.
Calculate the mass of 12.046 × 10²³ molecules of CaO.
Solution:
Gram molecular mass of CaO = 40 + 16 = 56 g
Mass of CaO = \(\frac{56 \times 12.046 \times 10^{23}}{6.023 \times 10^{23}}\)
56 × 2 = 112 g.

Question 27.
Calculate the number of moles for a substance containing 3.0115 × 10²³ molecules in it.
Solution:
Number of moles = \(\frac{\text { No. of molecules }}{\text { Avogadro number }}\)
= \(\frac{12.046 \times 10^{22}}{6.023 \times 10^{23}}\)
= 0.5 mole.

Question 28.
Calculate the number of moles in 12.046 × 1022 atoms of copper.
Solution:
No. of moles of atoms
\(\begin{array}{l}{=\frac{\text { No. of atoms }}{\text { Avogadro number }}} \\ {=\frac{12.046 \times 10^{22}}{6.023 \times 10^{23}}=2 \times 10^{-1}} \\ {=0.2 \text { mole. }}\end{array}\).

Question 29.
Calculate the number of moles in 24.092 × 10²² molecules of water.
Solution:
Number of moles =
\(\begin{array}{l}{=\frac{\text { No. of molecules }}{\text { Avogadro number }}} \\ {=\frac{24.092 \times 10^{22}}{6.023 \times 10^{23}}}\end{array}\)
= 4 × 1022 × 10^-23
= 4 × 10^-1
= 0.4 mole.

Question 30.
Which one of the following will have largest number of atoms?

1. 1 g Au (s)
2. 1 g Na (s)
3. 1 g Li (s)
4. 1 g of Cl₂ (g)
   (Atomic masses: Au = 197, Na = 23, Li = 7, Cl = 35.5 amu)

Solution:

1. 1 g Au = \(\frac{1}{197}\) mol = \(\frac{1}{197}\) × 6.02 × 10²³ atoms
2. 1 g Na = \(\frac{1}{23}\) mol = \(\frac{1}{23}\) × 6.02 × 10²³ atoms
3. 1 g Li = \(\frac{1}{7}\) mol = \(\frac{1}{7}\) × 6.02 × 10²³ atoms
4. 1 g Cl₂ = \(\frac{1}{71}\) mol = \(\frac{1}{71}\) × 6.02 × 10²³molecules = \(\frac{2}{71}\) × 6.02 × 10²³ atoms.

Thus, 1 g of Li has the largest number of atoms.

Question 31.
Calculate the number of atoms in each of the following:
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He
Solution:
(i) 1 mol of He = 6.022 × 10²³ atoms
52 mol of He = 52 × 6.022 × 10²³ atoms = 3.131 × 10²⁵ atoms.

(ii) 1 atom of He = 4 u of He
4 u of He = 1 atom of He
52 u of He = \(\frac{1}{4}\) × 52 atoms = 13 atoms.

(iii) 1 mole of He = 4 g = 6.022 × 10²³ atoms
52 g of He = \(\frac{6.022 \times 10^{23}}{4} \times 52\) atoms = 7.8286 × 1024 atoms.

Question 32.
Calculate the number of moles in each of the following.
(i) 392 g of sulphuric acid
(ii) 44.8 litres of sulphur dioxide at N.T.P.
(iii) 6.022 × 1022 molecules of oxygen
(iv) 8 g of calcium
Solution:
(i) 392 g of sulphuric acid
Molar mass of H₂SO₄ = 2 × 1 + 32 + 4 × 16 = 98 g
98 g of sulphuric acid = 1 mol
392 g of sulphuric acid = 1 mol × \(\frac{392 g}{(98 g)}\) = 4 mol.

(ii) 44.8 litres of sulphur dioxide at N.T.P.
22.4 litres of sulphur dioxide at N.T.P. = 1 mol
44.8 litres of sulphur dioxide at N.T.P. = \(\frac{1 \mathrm{mol}}{(22.4 \mathrm{L})} \times(44.8 \mathrm{L})\) = 2.0 mol.

(iii) 6.022 × 10²² molecules of oxygen
6.022 × 10²² molecules of oxygen = 1 mol
6.022 × 10²² molecules of oxygen = 1 mol × \(\frac{6.022 \times 10^{22}}{6.022 \times 10^{23}}\) = 0.1 mol.

(iv) 8 g of calcium
Gram atomic mass of Ca = 40 g
40 g of calcium = 1 mol
8.0 g of calcium = 1 mol × \(\frac{(8.0 \mathrm{g})}{(40 \mathrm{g})}\) = 0.2 mol.

Question 33.
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?
Solution:
Volume of a drop of water = 0.05 mL
Mass of a drop of water = Volume × density = (0.05 mL) × (1.0 g/mL) = 0.05 g
Gram molecular mass of water (H₂O) = 2 × 1 + 16 = 18 g
18 g of water = 1 mol
0.05 g of water = \(\frac{1 \mathrm{mol}}{(18 \mathrm{g})} \times(0.05 \mathrm{g})\) = 0.0028 mol
No. of molecules present
1 mole of water contain molecules = 6.022 × 10²³
0.0028 mole of water contain molecules = 6.022 × 1023 × 0.0028 = 1.68 × 10²¹ molecules.

Question 34.
Calculate the total number of electrons present in 1.6 g of methane.
Solution:
(i) Molar mass of methane (CH₄) = 12 + 4 × 1 = 16 g
16 g of methane contain molecules = 6.022 × 10²³
1.6 g of methane contain molecule = \(=\frac{6.022 \times 10^{23}}{(16 \mathrm{g})} \times(1.6 \mathrm{g})\) = 6.022 × 10²²

(ii) Number of electrons in 6.022 × 10²² molecules of methane
1 molecule of methane contains electrons = 6 + 4 = 10
6.022 × 1022 molecules of methane contain electrons = 6.022 × 1022 × 10 = 6.022 × 10²³

Question 35.
The Vapour Density of a gaseous element is 5 times that of oxygen under similar conditions. If the molecule is triatomic, what will be its atomic mass?
Solution:
Molecular mass of oxygen = 32 u
Density of oxygen = \(\frac{32}{2}\) = 16 u
Density of gaseous element = 16 × 5 = 80 u
Molecular mass of gaseous element = 80 × 2 = 160 u
Atomicity of the element = 3
Atomic mass of the element = \(\frac{\text { Molecular mass }}{\text { Atomicity }}=\frac{160}{3}\) = 53.33 u.

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.8

10th Maths Exercise 3.8 Samacheer Kalvi Question 1.
Find the square root of the following polynomials by division method
(i) x⁴ – 12x³ + 42x² – 36x + 9
(ii) 37x² – 28x³ + 4x⁴ + 42x + 9
(iii) 16x⁴ + 8x² + 1
(iv) 121x⁴ – 198x³ – 183x² + 216x + 144
Solution:
The long division method in finding the square root of a polynomial is useful when the degrees of a polynomial is higher.

Ex 3.8 Class 10 Samacheer Question 2.
Find the square root of the expression \(\frac{x^{2}}{y^{2}}-10 \frac{x}{y}+27-10 \frac{y}{x}+\frac{y^{2}}{x^{2}}\)
Solution:

10th Maths Exercise 3.8 Question 3.
Find the values of a and b if the following polynomials are perfect squares
(i) 4x⁴ – 12x³ + 37x² + bx + a
(ii) ax⁴ + bx³ + 361ax² + 220x + 100
Solution:
(i)

Since it is a perfect square.
Remainder = 0
⇒ b + 42 = 0, a – 49 = 0
b = -42, a = 49

(ii) ax⁴ + bx³ + 361ax² + 220x + 100

Since remainder is 0
a = 144
b = 264

Exercise 3.8 Class 10 Samacheer Question 4.
Find the values of m and n if the following expressions are perfect squares
(i) \(\frac{1}{x^{4}}-\frac{6}{x^{3}}+\frac{13}{x^{2}}+\frac{m}{x}+n\)
(ii) x⁴ – 8x³ + mx² + nx + 16
Solution:
(i)

(ii)

Since remainder is 0,
m = 24, n = -32

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.3

10th Maths Exercise 5.3 Samacheer Kalvi Question 1.
Find the equation of a straight line passing through the mid-point of a line segment joining the points(1, -5) (4, 2) and parallel to
(i) X axis
(ii) Y axis
Solution:

Exercise 5.3 Class 10 Samacheer Kalvi Question 2.
The equation of a straight line is 2(x – y) + 5 = 0. Find its slope, inclination and intercept on the Y axis.
Solution:
2(x – y) + 5 = 0
⇒ 2x – 2y + 5 =
⇒ 2y = 2x + 5

Ex 5.3 Class 10 Samacheer Question 3.
Find the equation of a line whose inclination is 30° and making an intercept – 3 on the Y axis.
Solution:
θ = 30°

10th Samacheer Maths Exercise 5.3 Answers Question 4.
Find the slope and y intercept of \(\sqrt{3}\)x + (1 – \(\sqrt{3}\))y = 3.
Solution:

10th Maths Exercise 5.3 Question 5.
Find the value of ‘a’, if the line through (-2, 3) and (8, 5) is perpendicular to y = ax = + 2
Solution:

10th Maths Coordinate Geometry Exercise 5.3 Question 6.
The hill in the form of a right triangle has its foot at (19, 3)The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Solution:
θ = 45°
Coordinate of foot of hill = (19, 3) let equation of line be y = mx + c
m = tan θ = tan 45° = 1
⇒ y = x + c
Substituting y = 3 & x = 19, 3 = 19 + c ⇒ c = -16

10th Maths 5.3 Question 7.
Find the equation of a line through the given pair of points

(ii) (2, 3) and (-7, -1)
Solution:
(i) Equation of the line in two point form is


⇒ 9y – 27 = 4x – 8
⇒ 4x – 9y – 8 + 27 = 0
⇒ 4x – 9y + 19 = 0

10th Standard Maths Exercise 5.3 Question 8.
A cat is located at the point(-6, -4) in xy plane. A bottle of milk is kept at (5, 11). The cat wish to consume the milk traveling through shortest possible distance. Find the equation of the path it needs to take its milk.
Solution:
A = (x₁, y₁) = (-6, -4)
B = (x₂, y₂) = (5, 11)
Shortest path between A and B is a line joining A and B.

10th Exercise 5.3 Question 9.
Find the equation of the median and altitude of ∆ABC through A where the vertices are A(6, 2) B(-5, -1) and C(1, 9)
Solution:

10th Maths 5.3 Solution Question 10.
Find the equation of a straight line which -5 has slope \(\frac{-5}{4}\) and passing through the point (-1, 2).
Solution:

10th Samacheer Kalvi Maths Coordinate Geometry Question 11.
You are downloading a song. The percent y (in decimal form) of mega bytes remaining to get downloaded in x seconds is given by y = -0.1x + 1.
(i) graph the equation.
(ii) find the total MB of the song.
(iii) after how many seconds will 75% of the song gets downloaded?
(iv) after how many seconds the song will be downloaded completely?
Solution:
(i) y = -0.1x + 1
when x = 0 ⇒ y = 1
when y = 0 ⇒ y = 10

(ii) Total MB of song can be obtained when time = 0
∴ x = 0
⇒ y = 1 MB
(iii) time when 75% of song is downloaded
⇒ remaining % = 25% ⇒ y = 0.25
0.25 = -0. 1x + 1
⇒ 0.1x = 0.75

(iv) song will downloaded completely when , remaining % = 0 ⇒ y = 0
⇒ 0 = -0.1x + 1
⇒ x = 10
∴ 10 seconds

10 Maths Exercise 5.3 Question 12.
Find the equation of a line whose intercepts on the x and y axes are given below.
(i) 4, -6
(ii) \(-5 \frac{3}{4}\)
Solution:

Maths Exercise 5.3 Question 13.
Find the intercepts made by the following lines on the coordinate axes,
(i) 3x – 2y – 6 = 0
(ii) 4x + 3y + 12 = 0
Solution:
(i) The given equation is
3x – 2y – 6 = 0
3x – 2y = 6
Divided by 6
\(\frac { 3x }{ 6 } \) – \(\frac { 2y }{ 6 } \) = \(\frac { 6 }{ 6 } \)
\(\frac { x }{ 2 } \) – \(\frac { y }{ 3 } \) = 1 ⇒ \(\frac { x }{ 2 } \) + \(\frac { y }{ -3 } \) = 1
(Comparing with \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1)
∴ x intercept = 2; y intercept = -3

(ii) The given equation is
4x + 3y + 12 = 0
4x + 3y = -12
Divided by -12
\(\frac { 4x }{ -12 } \) + \(\frac { 3y }{ -12 } \) = \(\frac { -12 }{ -12 } \)
\(\frac { x }{ -3 } \) + \(\frac { y }{ -4 } \) = 1
(Comparing with \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1)
∴ x intercept = -3; y intercept = -4

Ex 5.3 Class 10 Maths Solutions Question 14.
Find the equation of a straight line
(i) passing through (1, -4) and has intercepts which are in the ratio 2 : 5
(ii) passing through (-8, 4) and making equal intercepts on the coordinate axes
Solution:
(i) ratio of intercept = 2 : 5

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Tamilnadu Samacheer Kalvi 9th Social Science Geography Solutions Chapter 2 Lithosphere – II Exogenetic Processes

Lithosphere – II Exogenetic Processes Textual Exercise

I. Choose the correct answer.

Lithosphere 2 Exogenetic Processes Question 1.
The disintegration or decomposition of rocks is generally called as …………………
(a) weathering
(b) erosion
(c) transportation
(d) deposition
Answer:
(a) weathering

Lithosphere 2 Exogenic Processes Question 2.
The process of the levelling up of land by means of natural agents.
(a) aggradation
(b) degradation
(c) gradation
(d) none
Answer:
(c) gradation

Lithosphere 2 Exogenic Processes Book Back Answers Question 3.
……………. is seen in the lower course of the river.
(a) Rapids
(b) Alluvial fan
(c) Delta
(d) Gorges
Answer:
(c) Delta

9th Social Guide Question 4.
Karst topography is formed due to the action of ………………
(a) Glacier
(b) Wind
(c) Sea waves
(d) Ground water
Answer:
(d) Ground water

9th Std Social Science Guide Pdf English Medium Question 5.
Which one of the following is not a depositional feature of a glacier?
(a) cirque
(b) Moraines
(c) Drurrtlins
(d) Eskers
Answer:
(a) cirque

Chapter 2 Lithosphere – Ii Exogenetic Processes Question 6.
Deposits of fine silt blown by wind is called as …………….
(a) Loess
(b) Barchans
(c) Hamada
(d) Ripples
Answer:
(a) Loess

Samacheer Kalvi 9th Social Book Answers Question 7.
Stacks are formed by …………….
(a) Wave erosion
(b) River erosion
(c) Glacial erosion
(d) Wind deposion
Answer:
(a) Wave erosion

Social Guide For Class 9 Question 8.
……………… erosion is responsible for the formation of cirque.
(a) wind
(b) glacial
(c) river
(d) underground water
Answer:
(b) glacial

9th Std Social Science Guide Pdf Question 9.
Which one of the following is a second order land form?
(a) Asia
(b) Deccan Plateau
(c) Kulu valley
(d) Marina Beach
Answer:
(b) Deccan Plateau

II. Match the following.

Answer:
1. (c)
2. (d)
3. (a)
4. (e)
5. (b)

III. Answer in brief.

9th Social Science Guide Pdf Download Question 1.
Define weathering.
Answer:
Weathering is the disintegration and decomposition of materials of the earth’s crust by their exposure to the atmosphere.

Samacheer Kalvi 9th Social Book Solutions Question 2.
What do you mean by biological weathering?
Answer:
Biological weathering occurs due to the penetration and expansion of plant roofs, earth worms, burrowing animals (rabbits, rats) and same human activities.

9th Standard Social Book Question 3.
Mention the three courses of a river with any two land forms associated to each course.
Answer:
The course of a river is divided into

1. The Upper course – V shaped valleys, gorges
2. The middle course – Alluvial fans, flood plains
3. The lower course – Delta and Estuary.

Samacheer Kalvi 9th Social Book Question 4.
What are Ox-bow lakes?
Answer:

1. Meanders in due course of time become almost a complete circle with narrow necks.
2. This in turn gets abandoned and forms a lake. This is called an Ox-bow lake.

Class 9 Social Science Guide Question 5.
How does a cave differ from a sea arch?
Answer:

Sea cave	Sea Arch
Prolonged wave attack on the base of a cliff erodes rock materials which result in the formation of caves.	When two caves approach one another from either side of a headland and Unite, they form an arch, e.g., Neil Island, Andaman & Nicobar.

Samcheer Kalvi.Guru 9th Social Question 6.
List out any four karst topographical areas found in India.
Answer:

1. Guptadham caves – Western Bihar
2. Pandav caves Pachmuri – Madhya Pradesh
3. Kutumsar – Bastar district in Chattisgarh
4. Borra caves of Visakhapatnam – Andhra Pradesh

9th Geography Guide Question 7.
What do you mean by a hanging valley?
Answer:
These are valleys eroded by tributary glacier and that hangs over the main valley.

9th Standard Social Science Guide Pdf Question 8.
Define: (a) Moraine (b) Drumlin (c) Esker.
Answer:
(a) Moraine:

• Landforms formed by the glacial deposits of valley (or) continental glaciers are termed as moraines.
• They are of various shapes and sizes, like ground, terminal and lateral moraines, etc.

(b) Drumlin: Drumlins are deposits of glacial moraines that resemble giant inverted teaspoons (or) half cut eggs.
(c) Esker: Long narrow ridges composed of boulders gravel and sand deposited by streams of melting water which run parallel to a glacier are called eskers.

9th Std Social Science Question 9.
Mention the various features formed by wind erosion.
Answer:

• The erosional landforms of wind: Mushroom rocks, Inselbergs and Yardangs
• The Depositional Landforms of wind: Sand dunes, Barchans and loess.

Exogenetic Processes Part 2 9th Std Question 10.
What are wave cuLplatforms?
Answer:
Flat surface found at the foot of the sea cliffs are called as wave cut platforms, wave cut platform is also referred as beach, shelf, terrace and plain.

IV. Distinguish between:

Question 1.
Physical and chemical weathering
Answer:

S.No.	Physical weathering	Chemical weathering
(i)	It is the breakdown of rocks without changing their chemical composition through the action of physical forces.	Disintegration and. decomposition of rocks due to chemical reactions is called Chemical weathering.
(ii)	Cracks are formed and disintegration occurs eventually.	Chemical weathering takes place through the processes of oxidation, carbonation solution and hydration.
(iii)	Exfoliation, block disintegration, granular disintegration, etc., are the different types of weathering.	The agents of Chemical weathering are oxygen, CO2 and Hydrogen.

Question 2.
Delta and Estuary
Answer:

S.No.	Delta	Estuary
(i)	A triangular shaped low lying area formed by the river at its mouth is called Delta.	Deltas have fine deposits of sediments enriched with minerals.
(ii)	Estuary is formed where the rivers meet the sea.	Deposition of silt by the river is not possible in the estuaries like Delta as if the waves keep on eroding the deposits e.g. River Narmada, River Tapti.

Question 3.
Stalactite and stalagmite
Answer:

Stalactite	Stalagmite
When the water containing dissolved calcite gradually drips from the ceiling of the caves, water evaporates and the remaining calcite hangs from the ceiling and thus Stalactites are formed.	When the calcite deposits rises upward like a pillar Stalagmites are formed.

Question 4.
Longitudinal and Transverse sand dunes
Answer:

Longitudinal	Transverse sand dunes.
Longitudinal dunes are long narrow ridges of sand, which extend in a direction parallel to the prevailing winds. These dunes are called Seifs in Sahara.	Transverse dunes are asymmetrical in shape. They are formed by alternate slow and fast winds that blow from the same direction.

Question 5.
Inselbergs and yardangs
Answer:

S.No.	Inselbergs	Yardangs
(i)	Certain hard rocks like igneous rocks are more resistant to wind action.	In arid regions, certain rocks .have hard and soft layers arranged vertically.
(ii)	Isolated residual hills rising abruptly from their surroundings are termed as inselbergs. e.g., Uluru (or) Ayers Rock – Australia.	When winds blow over these rocks, the soft layers get eroded leaving irregular crests. These are called Yardangs.

Question 6.
Spit and bar
Answer:

Spit	Bar
A spit is a ridge (or) embankment of sediment, attached to the land on one end and terminating in open water on the other end. Spits are common at the mouth of estuaries, e.g., Kakinada Spit.	A bar is an elongated deposit of sand, shingle (or) mud found in the sea almost parallel to the shoreline.

V. Give Reasons.

Question 1.
Chemical weathering is predominant in hot and humid zones.
Answer:

1. Chemical is predominant in hot and humid zones because the warm temperature and rainfall increases the chemical weathering.
2. It encourages the decomposition of plant matter to produce chemicals such as humic acids and CO₂
   These chemicals increases the rate of weathering.

Question 2.
Slit deposits are less at estuaries than deltas.
Answer:
Silt deposits are less at estuaries than deltas because the waves keep on eroding the deposits e.g., River Narmada & Tapti.

Question 3.
The snow line is at the sea level in Polar regions.
Answer:
The snow line is at the sea level in Polar regions because higher the latitude lower the snow line from sea level. –

Question 4.
Wind can possibly erode the rocks from aO sides.
Answer:

• Wind erosion can occur in any area where the Soil (or) Sand is not compacted (or) if it is finely granulated in nature.
• Wind can loosen the materials and send it in all directions.

Question 5.
In limestone regions, surface drainage is rarely found.
Answer:

• The rain water mixes with CO₂ and enters into limestone region it dissolves and destroys much of the lime stone.
• Subterranean drainage may limit surface water with few to no rivers.
• The ability of water to penetrate the rock lowers. Joints and bedding planes make the rock permeable.
  So surface drainage is rarely found.

VI. Answer in Paragraph.

Question 1.
Write a note on weathering classify and explain.
Answer:
Weathering is the disintegration and decomposition of materials of the earth’s crust by their
exposure to atmosphere. There are three types of weathering,
(a) Physical weathering, (b) Chemical weathering (c) Biological weathering
Physical weathering: It is the breakdown of rocks without changing their chemical composition, through the action of physical forces. The constant freezing and thawing of rocks during the night and day leads to the expansion and contraction of rocks. Cracks are formed and disintegration occurs eventually. Exfoliation, block disintegration, granular disintegration etc., are the different types of weathering.

Chemical weathering: Disintegration and decomposition of rocks due to chemical reactions is called Chemical Weathering. This is predominantly high in the hot and humid regions such as the equatorial, tropical and sub tropical zones. Chemical weathering takes place through the processes of oxidation, carbonation, solution, and hydfation. The agents of Chemical weathering are Oxygen, Carbon-dioxide and Hydrogen.

Biological weathering: Biological weathering occurs due to the penetration and expansion of plant roots, earthworms, burrowing animals (rabbits, rats) and some human activities.

Question 2.
Explain the erosional landformsformed by underground water.
Answer:
Most erosion takes plaice due to the process of solution. When rain water mixes with carbon- di-oxide and enters into a limestone region, it dissolves and destroys much of the limestone. As a result, landforms suph as Terra rossa, Lappies, sinkholes, swallow holes, dolines, uvalas, poljes, caves and caverns are formed.

Terra Rossa (Italian term for Red soil): Deposition of red clay soil on the surface of the Earth is due to the dissolution of limestone content in rocks. The redness of the soil is due to the presence of iron oxide.

Lappies: When the joints of limestone rocks are corrugated by groundwater, long furrows are formed and these are called LAPPIES.

Sinkhole: A funnel shaped depressions formed due to dissolution of limestone rock is called sinkholes. Their average depth ranges between three and nine meters.

Caves and Caverns: Caves and caverns are subterranean features of karst topography. Caves are hollows that are formed by the dissolution of limestone rocks when carbon di oxide in air turns into carbonic acid after its reaction with water. They vary in size and shape. Caverns are the caves with irregular floors, e.g., Guptadham caves in Western Bihar.
All types of deposits in the caves and caverns are collectively called speleothems which includes travertines, tufa, dripstones.
Swallow Holes, Uvalas, Dolines, Poljis are other erossional Features of karst regions predominant in other parts of the world.

Question 3.
What is a glacier? Explain its types.
Answer:
A Glacier is a large mass of ice that moves slowly over the land, from its place of accumulation. It is also known as ‘River of ice’. The place of accumulation is called snowfield. The height above which there is a permanent snow cover in the higher altitude or latitude is called snowline. Higher the latitude, lower the snowline from sea level.

The gradual transformation of snow into granular ice is called ‘firn’ or ‘ neve’ and finally it becomes solid glacial ice.
Movement of Glacier: The large mass of ice creates pressure at its bottom and generates heat. Due to this, the glacier melts a little and starts to move The rate of movement of a glacier varies from a few centimetres to several hundred meters a day. The movement of glaciers depends on slope, volume of the glacier, thickness, roughness at the bottom (friction) etc., and Temperature. Like the rivers, glaciers also carry out erosion, transportation and deposition.

Types of Glacier: Glaciers are broadly divided into two types based on the place of occurrence, such as Continental glacier and valley glacier.

Question 4.
Describe the depositional work of winds.
Deposition occurs when the speed of wind is reduced by the presence of obstacles like bushes, forests and rock structures. The sediments carried by wind get deposited on both the wind ward and leeward sides of these obstacles.
Some of the depositional landforms are sand dunes, barchans and loess.

Sand Dune: In deserts, during sandstorms, wind carries loads of sand. When the speed of wind decreases, huge amount of sand gets deposited. These mounds or hills of sand are called sand dunes. There are different types of sand dunes.

Barchan: Barchans are isolated, crescent shaped sand dunes. They have gentle slopes on the windward side and steep slopes on the leeward side.
Transverse Dunes: Transverse dunes are asymmetrical in shape. They are formed by alternate slow and fast winds that blow from the same direction.

Longitudinal Dunes: Longitudinal dunes are long narrow ridges of sand, which extend in a direction parallel to the prevailing winds. These dunes are called Seifs in Sahara.

Loess: The term loess refers to the deposits of fine silt and porous sand over a vast region. Extensive loess deposits are found in Northern and Western China, the Pampas of Argentina, in Ukraine and in the Mississippi Valley of the United States.

Question 5.
Give a detailed account on the three orders of land forms.
Answer:
Major land forms:
(i) First order landforms : Continents & Oceans
(ii) Second order landforms : Mountains, Plateaus, and plains minor land forests
(iii) Third order landforms : Deltas, Fjords coasts, Sand dimes, Beaches, Valleys, Cirques, Mushroom rocks, Limestone rocks.

First order land forms:

1. Continents: (i) It is a very large area of land.
   (ii) One of the seven large land masses on the earth’s surface, surrounded by sea.
   Asia, Africa, Europe, North America, South America, Australia and Antarctica.
2. Oceans: A very large expanse of sea, Atlantic ocean, Arctic ocean, Pacific ocean, Indian ocean and Antarctic ocean.

Second order land forms: Mountains, Plateaus and Plains.

1. Mountains: A large natural elevation of the earth’s surface, rising abruptly from the surrounding level, e.g., Himalayas.
2. Plateaus: An area of fairly level high ground, e.g., Tibetean plateau.
3. Plains: A large area of flat land, e.g., Coastal plains.

Third order land forms: Deltas, Fjords, Sand dunes, Beaches, Valleys, Cirques, Mushroom rocks, Limestone rocks.

1. Deltas: A triangular shaped low lying area formed by the river at its mouth is called Delta. Fjords: These are glacial valleys that are partly submerged in the sea.
2. Sand Dunes: In deserts, huge amount of sand gets deposited. These mounds (or) hills of sand are called sand dunes.
3. Beaches: Sand and gravel are moved and deposited by waves along the shore to form Beaches. Valleys: A low area of land between hills (or) mountains typically with a river (or) stream flowing through it.
4. Cirques: The glacier erodes the steep side walls of the mountain and farms bowl shaped arm chair. It is termed as cirque.
5. Mushroom rocks: By the constant wearing down action of wind the bottom of the rock gets eroded away to form a mushroom like structure. ,
   This is called Mushroom rock (or) Pedestal rock.
6. Limestone rocks: The underground water creates distinct landforms in limestone regions called Karst Topography. It consists of calcite, argonite.

VII. Consider the given statements and choose the right option given below.

Question (i).
1. ‘I’ shaped valley is an erosional feature of the river.
2. ‘U’ shaped valley is an erosional feature of the glacier.
3. ‘V’ shaped valley is an erosional feature of the glacier.
(a) (i), (ii) and (iii) are right
(b) (i) and (ii) are right
(c) (i) and (iii) are right
(d) only (i) is right
Answer:
(d) only (ii) is right

Question (ii).
Statement I: Running water is an important agent of gradation.
Statement II: The work of the river depends on the slope of land on which-it flows.
(a) Statement I is false II is true
(b) Statement I and II are false
(c) Statement I is true II is false
(d) Statement I and II are true
Answer:
(a) Statement I is false II is true

Question (iii).
Statement: Limestone regions have less underground water.
Reason : Water does not percolate through limestone.
(a) The statement is right reason is wrong.
(b) The statement is wrong Reason is right.
(c) The statement and reason are wrong.
(d) The statement and reason are right.
Answer:
(d) The statement and reason are right.

VIII. HOTS

Question 1.
Is wind the only gradational agent in the desert?
Answer:
Yes, wind is the only gradational agent in the desert.
e.g., Erosional activity: Yardung
Depositional activity: Sand Dimes.

Question 2.
Underground water is more common in limestone areas than surface run off. Why?
Answer:
The chief constituent of limestone is calcium carbonate which is soluble in pure water and easily soluble in carbonate water.

Question 3.
The river channels in the lower course are wider than the upper course.
Answer:
The reasons are,

1. The river splits into a number of channels called distributaries.
2. The river brings down loads of debris from its upper and middle.
3. The river deposits and develop typical landforms like Delta and Estuary.

In-text HOTs Questions

Question 1.
Is weathering a pre-requisite in the formation of soil?
Answer:

1. Yes, weathering a pre-requisite in the formation of soil.
2. The rock materials in due course of time are weathered further to form soil.
3. Soil is a mixture of disintegrated rock material.

Question 2.
Snowline of Alps’is 2700 metre where as the snowline of Greenland is just 600 mts. Find out the reason.
Answer:
On tropical mountains the snowline may be as high as 500 mts, but when traced poleward it descends to 2700 mts in the European Alps to 600 meters in Greenland and just to se-a level near the poles. –

IX. Map Skill.

Question 1.
On the given outline map of the world, mark the following.
1. Any two deltas
2. A Karst region
3. Any two hot and cold deserts
Answer:
1. Any two deltas – Euphrates & Tigris Delta and Amazon river Delta

2. A Karst region – China

3. Any two hot and cold desets

X. Give geographical terms for the following:

Question 1.
(a) Chemical alternation of carbonate rocks on lime stone region.
(b) Flat surfaces near cliffs.
(c) Erosion + Transportation + Deposition =
(d) The bottom line of a snow field.
Answers:
(a) Carbonation
(b) Plateau
(c) Gradation
(d) The snowline

Lithosphere – I Endogenetic Processes Additional Questions

I. Choose the correct answer.

Question 1.
The process of dissolution of rock substances in water is …………….
(a) oxidation
(b) solution
(c) gradation
(d) hydration
Answer:
(b) solution

Question 2.
This generally originate from mountains.
(a) lake
(b) sea
(c) river
(d) ocean
Answer:
(c) river

Question 3.
The cylindrical holes drilled vertically in the river bed are …………….
(a) Pot holes
(b) canyons
(c) rapid
(d) Gorge
Answer:
(a) Pot holes

Question 4.
The largest Delta in the world is …………….
(a) The Nile River Delta
(b) The Ganga – Brahmaputra Delta
(c) The Yellow river Delta
(d) The Indus Delta
Answer:
(b) The Ganga – Brahmaputra Delta

Question 5.
The redness of the red clay soil is due to. the presence of ……………
(a) iron oxide
(b) carbon
(c) copper
(d) magnesium
Answer:
(a) iron oxide

Question 6.
The most powerful agents of gradation are ……………
(a) Rivers
(b) Glaciers
(c) Sea waves
(d) Streams
Answer:
(c) Sea waves

II. Match the following.

Answer:
1. (e)
2. (a)
3. (d)
4. (b)
5. (a)

III. Answer in brief.

Question 1.
Define Granular Disintegration.
Answer:
Granular disintegration takes place in crystalline rocks where the grains of the rocks become loose and fall out. This is due to the action of temperature & frost.

Question 2.
Mention the land features carved by a river in its upper course.
Answer:
‘V’ shaped valleys, Gorges, Canyons, rapids, pot holes, spurs and waterfalls.

Question 3.
What are “Pot Holes”?
Answer:
Due to the river action, cylindrical holes are drilled vertically in the river bed with varying depth and diametre.

Question 4.
State the other erosional features of Karst regions in other parts of the world.
Answer:
Swallow Holes, Uvalas, Dolines, Poljis are the other erosional features.

Question 5.
What are Transverse Dunes?
Answer:

1. Transverse Dunes are asymmetrical in shape.
2. They are formed by alternate slow and fast winds that blow from the same direction.

Question 6.
What are wave cut Platforms?
Answer:
Flat surface found at the foot of the sea cliffs are called as wave Cut platform. It is also referred as Beach, shelf, terrace and plain.

IV. Distinguish between.

Question 1.
Oxidation and Carbonation.
Answer:

S.No.	Oxidation	Carbonation
(i)	Oxygen in the atmosphere reacts with the Iron found in rocks thus leading to the formation of Iron oxide. This process is known as oxidation.	Carbonation is the mixing of water with the atmospheric CO2 forming carbonic acid.
(ii)	Oxidation weakens the rocks.	It is important in the formation of caves, in limestone region.

Question 2.
Alluvial Plain and Flood Plain.
Answer:

S.No.	Alluvial Plain	Flood Plain
(i)	A fan shaped deposition made by the river at the foothills is called an alluvial plain.	Fine sediments are deposited on river banks when a river floods and is called flood plain.
(ii)	These deposits are rich and fertile useful for cultivation.	These sediments make the region rich and fertile.

Question 3.
Arete and Matterhorn.
Answer:

Arete	Matterhorn
Aretes are narrow ridges formed when two cirque walls joined together back to back and forms narrow knife like ridges.	The pyramidal peaks formed when three (or) more cirques meet together are referred as matterhoms.

Question 4.
Sea Cave and Sea Arch.
Answer:

Sea Cave	Arch
Prolonged wave attack on the base of a cliff erodes rock materials which result in the formation of caves.	When two caves approach one another from either side of a headland and Unite, they form an arch, e.g., Neil Island, Andaman Nicobar.

V. Give reasons.

Question 1.
Why do the Biological weathering occur?
Answer:
Biological weathering occurs due to the penetration and expansion of plant roots, earth worms, burrowing animals (rabbits and rats) and some human activities.

Question 2.
Why is Karst Topography formed?
Answer:
Karst Topography is formed due to the dissolution of soluble rocks such as limestone, dolomite and Gypsum.

Question 3.
Why do the Pedestal rock look like mushroom?
Answer:
By the constant wearing down action of wind, the bottom gets eroded away to form a mushroom like structure. So the Pedestal rock looks like mushroom.

VI. Answer in a Paragraph.

Question 1.
Explain the origin of the river and its course.
Answer:
Rivers generally originate from mountains and end in a sea or lake. The whole path that a river flows through is called its course. The course of a river is divided into:
(i) The upper course
(ii) The middle course and
(iii) The lower course
(i) The Upper Course: Erosion is the most dominant action of river in the upper course. In this course, a river usually tumbles down the steep mountain slopes. The steep gradient increases the velocity and the river channel performs erosion with great force to widen and deepen its valley. The land features carved by a river in its upper course are V-shaped valleys, gorges, canyons, rapids, pot holes, spurs, and waterfalls.

(ii) The Middle Course: The river enters the plain in its middle course. The volume of water increases with the confluence of many tributaries and thus increases the load of the river. Thus, the predominant action of a river is transportation. The deposition also occurs due to the sudden decrease in velocity. The river in the middle course develops some typical landforms like alluvial fans, flood plains, meanders, ox-bow lakes etc.,

(iii) The Lower course: The river, moving downstream across a broad, level plain is loaded with debris, brought down from its upper and middle courses. Large deposits of sediments are found at the level bed and the river, splits into a number of channels called distributaries. The main work of the river here is deposition and it develops typical landforms like delta and estuary.

Question 2.
Describe the Erosional landforms of Sea.
Answer:
Some of the erosional landforms of sea waves are sea cliff, sea cave, arch, stack, beach, bar and spit and wave cut platform.

• Sea Cave: Prolonged wave attack on the base of a cliff erodes rock materials, which result in the formation of caves.
• Sea Arch: When two caves approach one another from either side of a headland and unite, they form an arch, e.g., Neil Island, Andaman and Nicobar.
• Sea Stack: Further erosion by waves ultimately leads to the total collapse of the arch. The seaward portion of the headland will remain as a pillar of rock known as stack. Eg the Old man of Hoy in Scotland.
• Sea Cliffs: Sea cliffs are steep rock faces formed when sea waves dash against them. The rocks get eroded to form steep vertical walls.
• Wave Cut Platforms: Flat surface found at the foot of sea cliffs are called as wave cut platforms. Wave cut platform is also referred as beach, shelf, terrace and plain.

VII. Consider the given statements and choose the right option given below.

Question 1.
(i) The nature and magnitude of weathering differs from place to place and region to region.
(ii) Granular disintegration takes place due to the action of volcanoes.
(iii) Weathering is a pre-requisite in the formation of soil.
Which of the above statement is/are the right statement.
(a) (i), (ii) and (iii) are right
(b) (i) & (ii) are right
(c) (i) & (iii) are right
(d) only (i) is right.
Answer:
(c) is right

Question 2.
(i) Small streams that join the main river is tributary.
(ii) River Gangas is a tributary.
Which of the above statement is/are the right statement.
(a) The statement is right reason is wrong
(b) The statement is wrong reason is right
(c) The statement & reason are wrong
(d) The statement & reason are right
Answer:
(a) is right.

VIII. Map Skill.

Question 1.
Indus and Ganga Brahmaputra Delta
Answer:

2. Sri Lanka & Myanmar

Students can Download Tamil Chapter 1.1 கலங்கரை விளக்கம் 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 2 Chapter 1.1 கலங்கரை விளக்கம்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.
Question 1.
வேயாமாடம் எனப்படுவது ………..
அ) வைக்கோலால் வேயப்படுவது
ஆ) சாந்தினால் பூசப்படுவது
இ) ஓலையால் வேயப்படுவது
ஈ) துணியால் மூடப்படுவது
Answer:
ஆ) சாந்தினால் பூசப்படுவது

Question 2.
உரவுநீர் அழுவம் – இத்தொடரில் அடிக்கோடிட்ட சொல்லின் பொருள்
அ) காற்று
ஆ) வானம்
இ) கடல்
ஈ) மலை
Answer:
இ) கடல்

Question 3.
கடலில் துறை அறியாமல் கலங்குவன …….
அ) மீன்கள்
ஆ) மரக்கலங்கள்
இ) தூண்க ள்
ஈ) மாடங்கள்
Answer:
ஆ) மரக்கலங்கள்

Question 4.
தூண் என்னும் பொருள் தரும் சொல் …..
அ) ஞெகிழி
ஆ) சென்னி
இ) ஏணி
ஈ) மதலை
Answer:
ஈ) மதலை

குறுவினா

Question 1.
மரக்கலங்களைத் துறை நோக்கி அழைப்பது எது?
Answer:

• மரக்கலங்களைத் துறை நோக்கி அழைப்பது கலங்கரை விளக்கம்.
• கடலில் துறை அறியாமல் கலங்கும் மரக்கலங்களைத் தன்னை நோக்கி கலங்கரை விளக்கம் அழைப்பதாகக் கடியலூர் உருத்திரங்கண்ணனார் தம் பாடலில் அழகாக சிறப்பித்துக் கூறுகிறார்.

Question 2.
கலங்கரை விளக்கில் எந்நேரத்தில் விளக்கு ஏற்றப்படும்?
Answer:
கலங்கரை விளக்கில் இரவு நேரத்தில் விளக்கு ஏற்றப்படும்.

சிறுவினா

கலங்கரை விளக்கம் பற்றிப் பெரும்பாணாற்றுப்படை கூறும் கருத்துகளை எழுதுக.
Answer:

• கலங்கரை விளக்கமானது வானம் கீழே விழாமல் தாங்கிப் பிடிக்கும் தூண் போலத் தோற்றமளிக்கிறது.
• ஏணி கொண்டு ஏற முடியாத அளவிற்கு உயரமாக உள்ளது.
• திண்மையான அதாவது திடமான கலவைச் சாந்து கொண்டு பூசப்பட்ட வானத்தை முட்டும் மாடத்தை உடையது.
• இந்த கலங்கரை விளக்கத்தில் இரவில் ஏற்றப்பட்ட எரியும் விளக்கு , திசை தெரியாமல் தவிக்கும் மரக்கலங்களைத் தன் துறை நோக்கி அழைப்பதாக புலவர்தம் பாடலில் குறிப்பிட்டுள்ளார்.

சிந்தனை வினா

கலங்கரை விளக்கம் கப்பல் ஓட்டிகளைத் தவிர வேறு யாருக்கெல்லாம் பயன்படும் என நீங்கள் கருதுகிறீர்கள்?
Answer:

• கடலில் சென்று மீன் பிடிக்கும் மீனவர்கள் மீன் பிடித்து திரும்பவும் கரை சேரவும் இக்கலங்கரை விளக்கம் பெரிதும் துணை புரிகின்றது.
• கடற்பயணம் சென்று கரை திரும்பும் பயணிகளுக்கு இக்கலங்கரை விளக்கம் சிறந்த வழிகாட்டியாக விளங்குகிறது.

கற்பவை கற்றபின்

Question 1.

கடற்கரைக்குச் சென்று அங்குள்ள காட்சிகளைக் கண்டு மகிழ்க.
Answer:

• நம்முடைய விடுமுறையை மகிழ்ச்சியாக கழிப்பதற்காகத் தேர்ந்தெடுக்கும் ஒரு முக்கிய இடம் கடற்கரை.
• பல இயற்கைச் சூழல்கள் கலந்த நம்முடைய நாட்டில் பல மாநிலங்கள் கடற்கரையோடு ஒட்டியுள்ளன.

மெரினா கடற்கரை :

• தமிழ்நாட்டின் தலைநகரான சென்னையில் அமைந்துள்ள மெரினா கடற்கரை உலகின் நீளமான கடற்கரைகளில் ஒன்றாகும். இதன் நீளம் 13 கிலோ மீட்டராகும்.
• தமிழகக் கடற்கரைகளிலேயே எல்லா நாள்களிலும் அதிக அளவு மக்கள் கூடும் கடற்கரையாக மெரினா கடற்கரை உள்ளது.
• இதன் அருகில் அண்ணா நினைவிடம், எம்.ஜி.ஆர் நினைவிடம், ஜெயலலிதா நினைவிடம், கலைஞர் கருணாநிதி நினைவிடம் ஆகியவை உள்ளன.

கன்னியாகுமரி கடற்கரை :

• இந்தியாவின் தென்கோடியில் உள்ள கன்னியாகுமரியில் இந்தக் கடற்கரை அமைந்துள்ளது.
• உள்ளூர் மட்டுமல்லாது உலகிலுள்ள வெளிநாட்டுப் பயணிகளும் வந்து செல்லும் கடற்கரை இது.
• இங்கு வங்காள விரிகுடா, அரபிக்கடல், இந்தியப் பெருங்கடல் இணையும் முக்கடல் சங்கமம் உள்ளது.
• கடலுக்குள் பாறையில் அமைந்திருக்கும் விவேகானந்தர் சிலை மற்றும் 133 அடி 6 உயரத்தில் அமைந்துள்ள திருவள்ளுவர் சிலை ஆகியவை மிகவும் புகழ்பெற்றவை.
• இங்கு சூரிய உதயத்தையும், சூரிய அஸ்தமனத்தையும் காண தினமும் ஆயிரக்கணக்கானவர்கள் வருகின்றனர்.

Question 2.
‘கலங்கரை விளக்கம்’ – மாதிரி ஒன்று செய்து வருக.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டியவை.

Question 3.
கடலும் கலங்கரை விளக்கமும் – ஓவியம் வரைந்து வண்ண ம் தீட்டுக.
Answer:

கூடுதல் வினாக்கள்

சொல்லும் பொருளும் :

1. மதலை – தூண்
2. சென்னி – உச்சி
3. ஞெகிழி – தீச்சுடர்
4. உரவுநீர் – பெருநீர்ப் பரப்பு
5. அழுவம் – கடல்
6. கரையும் – அழைக்கும்
7. வேயா மாடம் – வைக்கோல் போன்றவற்றால் வேயப்படாது. திண்மையாகச் சாந்து பூசப்பட்ட மாடம்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
‘மதலை’ என்னும் சொல்லுக்குப் பொருள் ………….
அ) தூண்
ஆ) உச்சி
இ) தீச்சுடர்
ஈ) கடல்
Answer:
அ) தூண்

Question 2.
‘ஞெகிழி’ என்னும் சொல்லுக்குப் பொருள்.
அ) தீச்சுடர்
ஆ) பெருநீர்ப் பரப்பு
இ) உச்சி
ஈ) அழைக்கும்
Answer:
அ) தீச்சுடர்

Question 3.
‘அழுவம்’ என்னும் சொல்லுக்குப் பொருள் ……..
அ) அழைக்கும்
ஆ) தூண்
இ) கடல்
ஈ) தீச்சுடர்
Answer:
இ) கடல்

Question 4.
‘சென்னி’ என்னும் சொல்லுக்குப் பொருள் …………
அ) உச்சி
ஆ) தீச்சுடர்
இ) அழைக்கும்
ஈ) கடல்
Answer:
அ) உச்சி]

Question 5.
‘உரவுநீர்’ என்னும் சொல்லுக்குப் பொருள்
அ) பெருநீர்ப் பரப்பு
ஆ) தீச்சுடர்
இ) உச்சி
ஈ) தூண்
Answer:
அ) பெருநீர்ப் பரப்பு

Question 6.
‘கரையும்’ என்னும் சொல்லுக்குப் பொருள் ………
அ) அழைக்கும்
ஆ) கடல்
இ) தீச்சுடர்
ஈ) உச்சி
Answer:
அ) அழைக்கும்

விடையளி:

Question 1.
கலங்கரை விளக்கம் என்றால் என்ன?
Answer:

• தமிழர் தம்முடைய மதிநுட்பத்தால் தொழில்நுட்ப அறிவினைக் கொண்டு கலம் படைத்து அதாவது, முதலில் படகைத் தயாரித்தனர். படிப்படியாக பாய்மரக்கப்பல், கப்பல் ஆகியவற்றை உருவாக்கினர்.
• படகுகளைக் கொண்டு மீன் பிடித்தும் வணிகம் செய்தும் வாழ்ந்து வந்தனர்.
• கடற்பயணம் சென்று கரை திரும்புவதற்குத் தமிழர் கண்ட தொழில்நுட்பமே கலங்கரை விளக்கமாகும்.

Question 2.
பத்துப்பாட்டு நூல்களைப் பட்டியலிடுக.
Answer:

•  திருமுருகாற்றுப்படை
• பொருநராற்றுப்படை
• பெரும்பாணாற்றுப்படை
• சிறுபாணாற்றுப்படை
• முல்லைப்பாட்டு
• மதுரைக்காஞ்சி
• நெடுநல்வாடை
• குறிஞ்சிப்பாட்டு
• பட்டினப்பாலை
• மலைபடுகடாம்

Question 3.
பெரும்பாணாற்றுப்படையின் நூலாசிரியர் பெயர் என்ன?
Answer:
பெரும்பாணாற்றுப்படையின் நூலாசிரியர் கடியலூர் உருத்திரங்கண்ணனார்.

Question 4.
கடியலூர் உருத்திரங்கண்ண னார் சங்கப் புலவரா?
Answer:
ஆம். இவர் சங்கப் புலவர்தான். இவர் கடியலூர் என்ற ஊரில் வாழ்ந்தவர்.

Question 5.
கடியலூர் உருத்திரங்கண்ணனார் இயற்றிய நூல்கள் யாவை?
Answer:

• பெரும்பாணாற்றுப்படை
• பட்டினப்பாலை

Question 6.
பெரும்பாணாற்றுப்படையின் பாட்டுடைத் தலைவன் பெயர் என்ன?
Answer:
தொண்டைமான் இளந்திரையன்.

Question 7.
ஆற்றுப்படை என்றால் என்ன?
Answer:
வள்ளல் ஒருவரிடம் பரிசு பெற்றுத் திரும்பும் புலவர், பாணர் போன்றோர் அந்த வள்ளலிடம் சென்று பரிசு பெற, பிறருக்கு வழிகாட்டுவதாகப் பாடப்படுவது ஆற்றுப்படையாகும்.

Question 8.
பத்துப்பாட்டில் இடம் பெறும் ஆற்றுப்படை நூல்களைப் பட்டியலிடுக.
Answer:

• திருமுருகாற்றுப்படை
• பொருநராற்றுப்படை
• பெரும்பாணாற்றுப்படை
• சிறுபாணாற்றுப்படை

பாடலின் பொருள்

கலங்கரை விளக்கமானது வானம் கீழே விழுந்துவிடாமல் தாங்கிக் கொண்டிருக்கும் தூண் போலத் தோற்றமளிக்கிறது; ஏணி கொண்டு ஏறமுடியாத உயரத்தைக் கொண்டிருக்கிறது; வேயப்படாமல் சாந்து பூசப்பட்ட விண்ணை முட்டும் மாடத்தை உடையது. அம்மாடத்தில் இரவில் ஏற்றப்பட்ட எரியும் விளக்கு, கடலில் துறை அறியாமல் கலங்கும் மரக்கலங்களைத் தன் துறை நோக்கி அழைக்கிறது.