Courses

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

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

The document NCERT Solutions of Atoms & Molecules (Class - 9 Science) Notes - Class 9 is a part of Class 9 category.
All you need of Class 9 at this link: Class 9
``` 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
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
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.

```
 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code

### Top Courses for Class 9

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;