NCERT Solutions of Atoms & Molecules (Class - 9 Science) Notes - Class 9

Class 9: NCERT Solutions of Atoms & Molecules (Class - 9 Science) Notes - Class 9

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 Page 1


CHAPTER- 3 ATOMS AND MOLECULES 
NCERT Solutions of Science  
Textbook Chapter Exercise Questions Solved  
Q.1: A 0.24 gm sample of compound of oxygen and boron was found by analysis to 
contain 0.096 gm of boron and 0.144 gm of oxygen. Calculate the percentage 
composition of the compound by weight.  
Ans: % of boron in sample = (0.096 ÷ 0.24) × 100 = 40% 
        % of oxygen in sample = (0.144 ÷ 0.24) × 100 = 60% 
       The sample of compound contains 40% boron and 60% oxygen by weight.  
Q.2: When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is 
produced. What mass of carbon dioxide will be formed when 3.0 gm of carbon is 
burnt in 50.00 gm of oxygen ? Which law of chemical combination will govern your 
answer ?  
Ans: When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is 
produced. It means all of carbon and oxygen are combined in the ratio of 3 : 8 to 
form carbon dioxide. Thus when there is 3 gm carbon and 50 gm oxygen, then also 
only 8 gm of oxygen will be used and 11 gm of carbon dioxide will be formed. The 
remaining oxygen is not used. 
This indicates law of definite proportions which says that in compounds, the 
combining elements are present in definite proportions by mass.  
Q.3: What are polyatomic ions ? Give examples. 
Ans: When two or more atoms combine together and behave like one entity with a 
net charge, then it is called polyatomic ion. For example, oxygen atom and hydrogen 
atom combine to form hydroxide ion (OH
–
). One carbon atom and three hydrogen 
atom combine to form carbonate ion (CO
3
–2
).  
Q.4: Write chemical formulae of the following:  
(a) Magnesium chloride (b) Calcium oxide (c) Calcium nitrate (d) Aluminium chloride 
(e) Calcium carbonate  
Ans: (a) MgCl
2
 (b) CaO (c) Cu(NO
3
)
2
 (d) AlCl
3
 (e) CaCO
3 
 
Q.5: Give the names of the elements present in the following compounds: 
(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate  
Ans: (a) Calcium and oxygen (b) Hydrogen and bromine (c) Sodium, hydrogen, 
carbon and oxygen (d) Potassium, sulphur and oxygen.  
Q.7: What is the mass of: 
(a) 1 mole of nitrogen atoms. 
(b) 4 moles of aluminium atoms. 
(c) 10 moles of sodium sulphite (Na
2
SO
3
).  
Ans:  
(a) 1 mole of nitrogen atoms = 14u = 14gm 
(b) 4 moles of aluminium atoms = 4 × 27 = 108u = 108gm 
(c) 1mole of sodium sulphite, Na
2
SO
3
 = 2 × 23 + 1 × 32 + 3 × 16 = 126u = 126gm 
Page 2


CHAPTER- 3 ATOMS AND MOLECULES 
NCERT Solutions of Science  
Textbook Chapter Exercise Questions Solved  
Q.1: A 0.24 gm sample of compound of oxygen and boron was found by analysis to 
contain 0.096 gm of boron and 0.144 gm of oxygen. Calculate the percentage 
composition of the compound by weight.  
Ans: % of boron in sample = (0.096 ÷ 0.24) × 100 = 40% 
        % of oxygen in sample = (0.144 ÷ 0.24) × 100 = 60% 
       The sample of compound contains 40% boron and 60% oxygen by weight.  
Q.2: When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is 
produced. What mass of carbon dioxide will be formed when 3.0 gm of carbon is 
burnt in 50.00 gm of oxygen ? Which law of chemical combination will govern your 
answer ?  
Ans: When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is 
produced. It means all of carbon and oxygen are combined in the ratio of 3 : 8 to 
form carbon dioxide. Thus when there is 3 gm carbon and 50 gm oxygen, then also 
only 8 gm of oxygen will be used and 11 gm of carbon dioxide will be formed. The 
remaining oxygen is not used. 
This indicates law of definite proportions which says that in compounds, the 
combining elements are present in definite proportions by mass.  
Q.3: What are polyatomic ions ? Give examples. 
Ans: When two or more atoms combine together and behave like one entity with a 
net charge, then it is called polyatomic ion. For example, oxygen atom and hydrogen 
atom combine to form hydroxide ion (OH
–
). One carbon atom and three hydrogen 
atom combine to form carbonate ion (CO
3
–2
).  
Q.4: Write chemical formulae of the following:  
(a) Magnesium chloride (b) Calcium oxide (c) Calcium nitrate (d) Aluminium chloride 
(e) Calcium carbonate  
Ans: (a) MgCl
2
 (b) CaO (c) Cu(NO
3
)
2
 (d) AlCl
3
 (e) CaCO
3 
 
Q.5: Give the names of the elements present in the following compounds: 
(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate  
Ans: (a) Calcium and oxygen (b) Hydrogen and bromine (c) Sodium, hydrogen, 
carbon and oxygen (d) Potassium, sulphur and oxygen.  
Q.7: What is the mass of: 
(a) 1 mole of nitrogen atoms. 
(b) 4 moles of aluminium atoms. 
(c) 10 moles of sodium sulphite (Na
2
SO
3
).  
Ans:  
(a) 1 mole of nitrogen atoms = 14u = 14gm 
(b) 4 moles of aluminium atoms = 4 × 27 = 108u = 108gm 
(c) 1mole of sodium sulphite, Na
2
SO
3
 = 2 × 23 + 1 × 32 + 3 × 16 = 126u = 126gm 
10 moles of sodium sulphite = 126 x 10 = 1260u = 1260gm.  
Q.8: Convert into mole 
(a) 12 gm of oxygen gas 
(b) 20 gm of water 
(c) 22 gm of carbon dioxide  
Ans: 
(a) 32 gm of oxygen gas = 1 mole 
12 gm of oxygen gas = 12 ÷ 32 = 0.375 mole. 
(b) 18 gm of water = 1 mole 
20 gm of water = 20 ÷ 18 = 1.1 mole. 
(c) 44 gm of carbon dioxide = 1 mole 
22 gm carbon dioxide = 22 ÷ 44 = 0.5 mole.  
Q.9: What is the mass of : 
(a) 0.2 mole of oxygen atoms 
(b) 0.5 mole of water molecules  
Ans:  
(a) 1 mole of oxygen atoms = 16 gm 
0.2 moles of oxygen atoms = 16 x 0.2 = 3.2 gm 
(b) 1 mole of water molecules = 18 gm 
0.5 mole of water molecules = 18 x 0.5 = 9 gm.  
Q.10: Calculate the number of molecules of sulphur (S
8
) present in 16 gm of solid 
sulphur. 
Ans: 1 mole of S
8
 = 32 x 8 = 256 gm 
1 mole of S
8
 = 6.023 x 10
23
 molecules 
So, 256 gm of S
8
 = 6.023 x 10
23 
S
8
 molecules 
Or, 16 gm S
8
 = (6.023 x 10
23 
÷ 256) x 16 = 3.76 x 10
22
 molecules.  
Q.11: Calculate the number of aluminium ions present in 0.51 gm of aluminium 
oxide. 
Ans: 1 mole of aluminium oxide, Al
2
O
3
 = 2 x 27 + 3 x 16 = 102u = 102 gm 
That is 102 gm Al
2
O
3 
has 6.023 x 10
23
 Al
2
O
3
 molecules 
Or, 0.51 gm Al
2
O
3 
has = (6.023 x 10
23
 x 0.51) ÷ 102 = 3.01 x 10
21
 Al
2
O
3
 molecules. 
We know 1 molecule of Al
2
O
3 
has 2Al
+++
 ions. 
Hence, 0.51 gm Al
2
O
3 
has = 2 x 3.01 x 10
21
 Al
+++
 ions  
= 6.023 x 10
21 
aluminium ions. 
 
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