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Page 1
NCERT
SOLUTIONS
CLASS - 9th
aglase .co
SOLUTIONS
CLASS - 9th
aglase .co
Page 2
Book : Science Ncert Solutions | Chapter-3 Science
Class : 9th
Subject : Science
Chapter : 3
Chapter Name : ATOMS AND MOLECULES
Q1 In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g
of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in
agreement with the law of conservation of mass.
sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water
Answer. In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium
ethanoate, carbon dioxide, and water.
Page : 32 , Block Name : Questions
Q2 Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen
gas would be required to react completely with 3 g of hydrogen gas?
Answer. It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8. Then, the
mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g. Therefore, the
mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 x 3 g = 24 g.
Page : 33 , Block Name : Questions
Q3 Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Page 1 of 8 Aglasem Schools
Class : 9th
Subject : Science
Chapter : 3
Chapter Name : ATOMS AND MOLECULES
Q1 In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g
of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in
agreement with the law of conservation of mass.
sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water
Answer. In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium
ethanoate, carbon dioxide, and water.
Page : 32 , Block Name : Questions
Q2 Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen
gas would be required to react completely with 3 g of hydrogen gas?
Answer. It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8. Then, the
mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g. Therefore, the
mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 x 3 g = 24 g.
Page : 33 , Block Name : Questions
Q3 Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Page 1 of 8 Aglasem Schools
Page 3
Book : Science Ncert Solutions | Chapter-3 Science
Answer.The postulate of Dalton's atomic theory which is a result of the law of conservation of
mass is: Atoms are indivisible particles, which can neither be created nor destroyed in a chemical
reaction.
Page : 33 , Block Name : Questions
Q4 Which postulate of Dalton’s atomic theory can explain the law of de nite proportions?
Answer. The postulate of Dalton's atomic theory which can explain the law of de nite proportion
is: The relative number and kind of atoms in a given compound remains constant.
Page : 33 , Block Name : Questions
Q1 . De ne the atomic mass unit?
Answer. Mass unit equal to exactly one-twelfth the mass of one atom of carbon-12 is called one
atomic mass unit. It is written as 'u'.
Page : 35 , Block Name : Questions
Q2 Why is it not possible to see an atom with naked eyes?
Answer.The size of an atom is so small that it is not possible to see it with naked eyes. Also, the
atom of an element does not exist independently.
Page : 35 , Block Name : Questions
Q1 Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer. (i) sodium oxide- Na O2
(ii) aluminium chloride-AlCl 3
(iii) sodium sulphide-Na S
2
(iv) magnesium hydroxide-Mg(OH) 2
Page : 39 , Block Name : Questions
Q2 Write down the names of compounds represented by the following formulae:
(i) Al (SO )
2 4 3
(ii) CaCl 2
(iii) K SO
2 4
Page 2 of 8 Aglasem Schools
Answer.The postulate of Dalton's atomic theory which is a result of the law of conservation of
mass is: Atoms are indivisible particles, which can neither be created nor destroyed in a chemical
reaction.
Page : 33 , Block Name : Questions
Q4 Which postulate of Dalton’s atomic theory can explain the law of de nite proportions?
Answer. The postulate of Dalton's atomic theory which can explain the law of de nite proportion
is: The relative number and kind of atoms in a given compound remains constant.
Page : 33 , Block Name : Questions
Q1 . De ne the atomic mass unit?
Answer. Mass unit equal to exactly one-twelfth the mass of one atom of carbon-12 is called one
atomic mass unit. It is written as 'u'.
Page : 35 , Block Name : Questions
Q2 Why is it not possible to see an atom with naked eyes?
Answer.The size of an atom is so small that it is not possible to see it with naked eyes. Also, the
atom of an element does not exist independently.
Page : 35 , Block Name : Questions
Q1 Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer. (i) sodium oxide- Na O2
(ii) aluminium chloride-AlCl 3
(iii) sodium sulphide-Na S
2
(iv) magnesium hydroxide-Mg(OH) 2
Page : 39 , Block Name : Questions
Q2 Write down the names of compounds represented by the following formulae:
(i) Al (SO )
2 4 3
(ii) CaCl 2
(iii) K SO
2 4
Page 2 of 8 Aglasem Schools
Page 4
Book : Science Ncert Solutions | Chapter-3 Science
(iv)KN O 3
(v) CaCO 3
Answer. (i) Al (SO ) -> Aluminium sulphate
2 4 3
(ii) CaCl -> Calcium chloride
2
(iii) K SO -> Potassium sulphate
2 4
(iv) KN O -> Potassium nitrate
3
(v) CaCO -> Calcium carbonate
3
Page : 39 , Block Name : Questions
Q3 What is meant by the term chemical formula?
Answer. The chemical formula of a compound means the symbolic representation of the
composition of a compound. From the chemical formula of a compound, we can know the number
and kinds of atoms of different elements that constitute the compound. For example, from the
chemical formula C02 of carbon dioxide, we come to know that one carbon atom and two oxygen
atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.
Page : 39 , Block Name : Questions
Q4 How many atoms are present in a
(i) H S molecule and
2
(ii) P O ion?3−
4
Answer. (i) In an H S molecule, three atoms are present; two of hydrogen and one of sulphur.
2
(ii) In a P O ion, ve atoms are present; one of phosphorus and four of oxygen.
3−
4
Q1 Calculate the molecular masses of H , O , Cl , CO , CH4, C2H6 ,C2H4 , NH3 , CH OH.
2 2 2 2 3
Answer.
Page 3 of 8 Aglasem Schools
(iv)KN O 3
(v) CaCO 3
Answer. (i) Al (SO ) -> Aluminium sulphate
2 4 3
(ii) CaCl -> Calcium chloride
2
(iii) K SO -> Potassium sulphate
2 4
(iv) KN O -> Potassium nitrate
3
(v) CaCO -> Calcium carbonate
3
Page : 39 , Block Name : Questions
Q3 What is meant by the term chemical formula?
Answer. The chemical formula of a compound means the symbolic representation of the
composition of a compound. From the chemical formula of a compound, we can know the number
and kinds of atoms of different elements that constitute the compound. For example, from the
chemical formula C02 of carbon dioxide, we come to know that one carbon atom and two oxygen
atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.
Page : 39 , Block Name : Questions
Q4 How many atoms are present in a
(i) H S molecule and
2
(ii) P O ion?3−
4
Answer. (i) In an H S molecule, three atoms are present; two of hydrogen and one of sulphur.
2
(ii) In a P O ion, ve atoms are present; one of phosphorus and four of oxygen.
3−
4
Q1 Calculate the molecular masses of H , O , Cl , CO , CH4, C2H6 ,C2H4 , NH3 , CH OH.
2 2 2 2 3
Answer.
Page 3 of 8 Aglasem Schools
Page 5
Book : Science Ncert Solutions | Chapter-3 Science
Page : 40 , Block Name : Questions
Q2 Calculate the formula unit masses of ZnO,Na O, K CO , given atomic masses of Zn = 65 u,
2 2 3
Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Answer.
Page 4 of 8 Aglasem Schools
Page : 40 , Block Name : Questions
Q2 Calculate the formula unit masses of ZnO,Na O, K CO , given atomic masses of Zn = 65 u,
2 2 3
Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Answer.
Page 4 of 8 Aglasem Schools
Page 6
Book : Science Ncert Solutions | Chapter-3 Science
Page : 40 , Block Name : Questions
Q1 If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?
Answer. One mole of carbon atoms weighs 12 g (Given)
i.e., mass of I mole of carbon atoms = 12 g
Then, mass of 6.022 × 10 number of carbon atoms = 12
23
Therefore, mass of I atom of carbon g
12
23
6.022×10
=1.9926 × 10 −21
g
Page : 42 , Block Name : Questions
Q2 Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic
mass of Na = 23 u, Fe = 56 u)?
Answer. Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains =6.022 × 10 number of atoms
23
23
Thus, 100 g of Na contains =
6.022×10
× 100
23
number of atoms= 2.6182 × 10 number of atoms,
24
Again, atomic mass of Fe = 56 u(Given)
Then, gram atomic mass of Fe = 56 g
23
Now, 56 g of Fe contains = number of atoms
6.022×10
× 100
56
Thus, 100 g of Fe contains =1.0753 × 10 numberof atoms
24
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of
iron.
Page : 42 , Block Name : Questions
Q1 A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of
boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer. Mass of boron= 0.096 g (Given)
Mass of oxygen = 0.144 (Given)
Mass of sample = 0.24 g (Given)
Thus, percentage of boron by weight in the compound = 0.096
0.24
× 100
=40 %
And, percentage of oxygen by weight in the compound =
0.144
× 100
0.24
=60%
Page : 43 , Block Name : Exercise
Q2 When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What
Page 5 of 8 Aglasem Schools
Page : 40 , Block Name : Questions
Q1 If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?
Answer. One mole of carbon atoms weighs 12 g (Given)
i.e., mass of I mole of carbon atoms = 12 g
Then, mass of 6.022 × 10 number of carbon atoms = 12
23
Therefore, mass of I atom of carbon g
12
23
6.022×10
=1.9926 × 10 −21
g
Page : 42 , Block Name : Questions
Q2 Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic
mass of Na = 23 u, Fe = 56 u)?
Answer. Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains =6.022 × 10 number of atoms
23
23
Thus, 100 g of Na contains =
6.022×10
× 100
23
number of atoms= 2.6182 × 10 number of atoms,
24
Again, atomic mass of Fe = 56 u(Given)
Then, gram atomic mass of Fe = 56 g
23
Now, 56 g of Fe contains = number of atoms
6.022×10
× 100
56
Thus, 100 g of Fe contains =1.0753 × 10 numberof atoms
24
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of
iron.
Page : 42 , Block Name : Questions
Q1 A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of
boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer. Mass of boron= 0.096 g (Given)
Mass of oxygen = 0.144 (Given)
Mass of sample = 0.24 g (Given)
Thus, percentage of boron by weight in the compound = 0.096
0.24
× 100
=40 %
And, percentage of oxygen by weight in the compound =
0.144
× 100
0.24
=60%
Page : 43 , Block Name : Exercise
Q2 When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What
Page 5 of 8 Aglasem Schools
Page 7
Book : Science Ncert Solutions | Chapter-3 Science
mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which
law of chemical combination will govern your answer?
Answer. Carbon + Carbon ->dioxide
3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide.
If 3 g of carbon is burnt In SO g of oxygen, then 3 g of carbon will react with 8 g of
oxygen. The remaining 42 g of oxygen will be left un-reactive.
In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
Page : 43 , Block Name : Exercise
Q3 What are polyatomic ions? Give examples.
Answer. A polyatomic ion is group of atoms carrying a charge (positive or negative). For
Example ammonium ion (NH ),hydroxide ion OH ,carbonate ion co ,sulphate ion SO
+ − −2 −2
4 3 4
Page : 44 , Block Name : Exercise
Q4 Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer.(a) Magnesium chloride- MgCl 2
(b) Calcium oxide - CaO
(c) Copper nitrate - Cu(NO ) 3 2
(d) Aluminium chloride - AlCl 3
(e) Calcium carbonate- CaCO 3
Page : 44 , Block Name : Exercise
Q5 Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer.
Page 6 of 8 Aglasem Schools
mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which
law of chemical combination will govern your answer?
Answer. Carbon + Carbon ->dioxide
3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide.
If 3 g of carbon is burnt In SO g of oxygen, then 3 g of carbon will react with 8 g of
oxygen. The remaining 42 g of oxygen will be left un-reactive.
In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
Page : 43 , Block Name : Exercise
Q3 What are polyatomic ions? Give examples.
Answer. A polyatomic ion is group of atoms carrying a charge (positive or negative). For
Example ammonium ion (NH ),hydroxide ion OH ,carbonate ion co ,sulphate ion SO
+ − −2 −2
4 3 4
Page : 44 , Block Name : Exercise
Q4 Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer.(a) Magnesium chloride- MgCl 2
(b) Calcium oxide - CaO
(c) Copper nitrate - Cu(NO ) 3 2
(d) Aluminium chloride - AlCl 3
(e) Calcium carbonate- CaCO 3
Page : 44 , Block Name : Exercise
Q5 Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer.
Page 6 of 8 Aglasem Schools
Page 8
Book : Science Ncert Solutions | Chapter-3 Science
Page : 44 , Block Name : Exercise
Q6 Calculate the molar mass of the following substances.
(a) Ethyne, C H
2 2
(b) Sulphur molecule, S 8
(c) Phosphorus molecule, P (Atomic mass of phosphorus = 31)
4
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO 3
Answer. (a) Molar mass of ethyne, C H = 2 x 12 + 2 x 1 = 26 g
2 2
(b) Molar mass of sulphur molecule, S = 8 x 32 =256 g
8
(c) Molar mass of phosphorus molecule,P = 4 x 31 = 124 g
4
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Molar mass of nitric acid, HNO = 1 + 14 + 3 x 16 = 63 g
3
Page : 44 , Block Name : Exercise
Q7 What is the mass of—
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3 )?
Answer. (a) The mass of 1 mole of nitrogen atoms is 14 g.
(b) The mass of 4 moles of aluminium atoms is (4 x 27) =108g
(c) The mass of 10 moles of sodium sulphite is 10 x 23 + 32+3 x 16] g = 10 x 126 g = 1260g
Page : 44 , Block Name : Exercise
Q8 Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Answer. (a) 32 g of oxygen gas =1 mole
then,12 g of oxygen gas=12/32 mole=0.375 moles
Page 7 of 8 Aglasem Schools
Page : 44 , Block Name : Exercise
Q6 Calculate the molar mass of the following substances.
(a) Ethyne, C H
2 2
(b) Sulphur molecule, S 8
(c) Phosphorus molecule, P (Atomic mass of phosphorus = 31)
4
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO 3
Answer. (a) Molar mass of ethyne, C H = 2 x 12 + 2 x 1 = 26 g
2 2
(b) Molar mass of sulphur molecule, S = 8 x 32 =256 g
8
(c) Molar mass of phosphorus molecule,P = 4 x 31 = 124 g
4
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Molar mass of nitric acid, HNO = 1 + 14 + 3 x 16 = 63 g
3
Page : 44 , Block Name : Exercise
Q7 What is the mass of—
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3 )?
Answer. (a) The mass of 1 mole of nitrogen atoms is 14 g.
(b) The mass of 4 moles of aluminium atoms is (4 x 27) =108g
(c) The mass of 10 moles of sodium sulphite is 10 x 23 + 32+3 x 16] g = 10 x 126 g = 1260g
Page : 44 , Block Name : Exercise
Q8 Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Answer. (a) 32 g of oxygen gas =1 mole
then,12 g of oxygen gas=12/32 mole=0.375 moles
Page 7 of 8 Aglasem Schools
Page 9
Book : Science Ncert Solutions | Chapter-3 Science
(b) 18 g of water = 1 mole
Then, 20 of water = 20/18 mole=1.11 moles (approx)
(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = 22/44 mole =0.5 mole
Page : 44 , Block Name : Exercise
Q9 What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer, (a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 16g
(b) Mass of one mole of water molecule = 18 g
Then, mass of 0.5 mole of water molecules = 0.5 x 18 g = 9 g
Page : 44 , Block Name : Exercise
Q10 Calculate the number of molecules of sulphur S present in 16 g of solid sulphur.
8
Answer. I mole of solid sulphur S = 8 x 32 g = 256 g
8
i.e., 256 g of solid sulphur contains = 6.022 × 10 molecules 23
21
Then, 16 g of solid sulphur contains = 6.022×10
256
× 16 molecules
=3.76 × 10 22
molecules
Page : 44 , Block Name : Exercise
Q11 Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
Answer. 1 mole of aluminium oxide( Al O )=2x27 +3 x 16 =102g
2 3
I.e.,102 gm of Al O =6.022 × 10 molecules of Al O
2 3
23
2 3
23
Then ,0.051 gm of Al O contains=
2 3
6.022×10
102
× 0.051 molecules
=3.011 × 10 molecules of Al O3
20
2
The number aluminium ions (Al present in one molecules of aluminium oxide is 2.
3+
)
Therefore,the number of aluminium ions (Al present in
3+
)
3.011 × 10
20
molecules of aluminium oxide =2 × 3.011 × 10 20
=6.022 × 10
20
Page : 44 , Block Name : Exercise
Page 8 of 8 Aglasem Schools
(b) 18 g of water = 1 mole
Then, 20 of water = 20/18 mole=1.11 moles (approx)
(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = 22/44 mole =0.5 mole
Page : 44 , Block Name : Exercise
Q9 What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer, (a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 16g
(b) Mass of one mole of water molecule = 18 g
Then, mass of 0.5 mole of water molecules = 0.5 x 18 g = 9 g
Page : 44 , Block Name : Exercise
Q10 Calculate the number of molecules of sulphur S present in 16 g of solid sulphur.
8
Answer. I mole of solid sulphur S = 8 x 32 g = 256 g
8
i.e., 256 g of solid sulphur contains = 6.022 × 10 molecules 23
21
Then, 16 g of solid sulphur contains = 6.022×10
256
× 16 molecules
=3.76 × 10 22
molecules
Page : 44 , Block Name : Exercise
Q11 Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
Answer. 1 mole of aluminium oxide( Al O )=2x27 +3 x 16 =102g
2 3
I.e.,102 gm of Al O =6.022 × 10 molecules of Al O
2 3
23
2 3
23
Then ,0.051 gm of Al O contains=
2 3
6.022×10
102
× 0.051 molecules
=3.011 × 10 molecules of Al O3
20
2
The number aluminium ions (Al present in one molecules of aluminium oxide is 2.
3+
)
Therefore,the number of aluminium ions (Al present in
3+
)
3.011 × 10
20
molecules of aluminium oxide =2 × 3.011 × 10 20
=6.022 × 10
20
Page : 44 , Block Name : Exercise
Page 8 of 8 Aglasem Schools
Document Details
| Board / Org | NCERT |
| Exam | Class 9 |
| Type | Solution |
| Pages | 9 |
| Updated | 22 Jul 2026 |