NCERT Solutions Class 9 Science Chapter 3 Atoms and Molecules

Class IX
Chapter 3 – Atoms and Molecules Science

Question 1: In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Answer: In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water.
Mass of sodium carbonate = 5.3 g (Given)
Mass of ethanoic acid = 6 g (Given)
Mass of sodium ethanoate = 8.2 g (Given)
Mass of carbon dioxide = 2.2 g (Given)
Mass of water = 0.9 g (Given)

Now, total mass before the reaction = (5.3 + 6) g
= 11.3 g

And, total mass after the reaction = (8.2 + 2.2 + 0.9) g
= 11.3 g

∴Total mass before the reaction = Total mass after the reaction
Hence, the given observations are in agreement with the law of conservation of mass.

Question 2: 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 × 3 g = 24 g.

Question 3: Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass? 
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.

Question 4: Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer: The postulate of Dalton’s atomic theory which can explain the law of definitevproportion is: The relative number and kind of atoms in a given compound remains constant.

Chapter 3 – Atoms and Molecules

Question 1: Define atomic mass unit.
Answer: Mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12 is called one atomic mass unit. It is written as ‘u’.

Question 2: Why is it not possible to see an atom with naked eyes?
Answer: Thesize 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.

Chapter 3 – Atoms and Molecules

Question 1: Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide
Answer:
(i) Sodium oxide →Na₂O
(ii) Aluminium chloride → AlCl₃
(iii) Sodium suphide → Na₂S
(iv) Magnesium hydroxide → Mg(OH)₂

Question 2: Write down the names of compounds represented by the following formulae:
(i) Al₂(SO₄)₃
(ii) CaCl₂ 
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃
Answer:
(i) Al₂(SO₄)₃ → Aluminium sulphate
(ii) CaCl₂ → Calcium chloride
(iii) K₂SO₄ → Potassium sulphate
(iv) KNO₃ → Potassium nitrate
(v) CaCO₃ → Calcium carbonate

Question 3: 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 CO2 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.

Question 4: How many atoms are present in a
(i) H₂S molecule and
(ii) PO₄³⁻ ion?
Answer: (i) In an H₂S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a PO₄³⁻ ion, five atoms are present; one of phosphorus and four of oxygen.

Chapter 3 – Atoms and Molecules

Question 1: Calculate the molecular masses of H₂, O2, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
Answer:
Molecular mass of H₂ = 2 × Atomic mass of H
= 2 × 1
= 2 u
Molecular mass of O₂ = 2 × Atomic mass of O
= 2 × 16
= 32 u
Molecular mass of Cl₂ = 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u
Molecular mass of CO₂ = Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16
= 44 u
Molecular mass of CH₄ = Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1
= 16 u
Molecular mass of C₂H₆ = 2 × Atomic mass of C + 6 × Atomic mass of H
= 2 × 12 + 6 × 1
= 30 u
Molecular mass of C₂H₄ = 2 × Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28 u
Molecular mass of NH₃ = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3 × 1
= 17 u 
Molecular mass of CH₃OH = Atomic mass of C + 4 × Atomic mass of H + Atomic mass of O
= 12 + 4 × 1 + 16
= 32 u
Question 2:
Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn
= 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Answer:
Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16
= 81 u
Formula unit mass of Na₂O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 + 16
= 62 u
Formula unit mass of K₂CO₃ = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 + 12 + 3 × 16
= 138 u

Chapter 3 – Atoms and Molecules

Question 1: If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
Answer:
One mole of carbon atoms weighs 12 g (Given)
i.e., mass of 1 mole of carbon atoms = 12 g
Then, mass of 6.022x10²³ number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon = 12/6.022x10²³g = 1.9926 x 10^-23g

Question 2: 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 6.022x10²³ = number of atoms
Thus, 100 g of Na contains 6.022x10²³/23 x 100 number of atoms
= 2.61826 x 10²⁴number of atoms
Again, atomic mass of Fe = 56 u(Given)
Then, gram atomic mass of Fe = 56 g
Now, 56 g of Fe contains 6.022x10²³= number of atoms
Thus, 100 g of Fe contains 6.022x10²³/56 x 100 number of atoms
= 1.0753 x 10²⁴number of atoms
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Chapter 3 – Atoms and Molecules

Question 1: 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 g (Given)
Mass of sample = 0.24 g (Given)
Thus, percentage of boron by weight in the compound = 0.096 /0.24 x 100% 
= 40%
And, percentage of oxygen by weight in the compound = 0.144/0.24 x 100%
= 60%

Question 2: When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What 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 combinations will govern your answer?
Answer: Carbon + Oxygen → 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 50 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.

Question 3: What are polyatomic ions? Give examples?
Answer: A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (NH⁺⁴) , hydroxide ion (OH−), carbonate ion (CO^2-₃) sulphate ion (SO^2-₄)

Question 4: 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 → MgCl₂
(b) Calcium oxide → CaO
(c) Copper nitrate → Cu (NO₃)₂
(d) Aluminium chloride → AlCl3
(e) Calcium carbonate → CaCO₃

Question 5: Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer: 

Question 6: Calculate the molar mass of the following substances:
(a) Ethyne, C₂H₂
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃
Answer:
(a) Molar mass of ethyne, C₂H₂ = 2 × 12 + 2 × 1 = 28 g
(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256 g
(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124 g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Molar mass of nitric acid, HNO₃ = 1 + 14 + 3 × 16 = 63 g

Question 7: 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 × 27) g = 108 g
(c) The mass of 10 moles of sodium sulphite (Na₂SO₃) is
10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

Question 8: 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 mole
(b) 18 g of water = 1 mole
Then, 20 g 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

Question 9: 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 = 3.2 g
(b) Mass of one mole of water molecule = 18 g
Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

Question 10: Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Answer:
1 mole of solid sulphur (S8) = 8 × 32 g = 256 g
i.e., 256 g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16 g of solid sulphur contains = 6.022 x 10²³ /256 x 16 molecules
= 3.76 × 10²² molecules (approx)

Question 11: Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element.
Atomic mass of Al = 27 u)
Answer:
1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102 g
i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3
Then, 0.051 g of Al₂O₃ contains =  6.022 x 10²³ /102 x 0.051 molecules
= 3.011 × 1020 molecules of Al₂O₃
The number of aluminium ions (Al³⁺) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al³⁺) present in 3.011 × 10²⁰ molecules
(0.051 g ) of aluminium oxide (Al2O3) = 2 × 3.011 × 10²⁰
= 6.022 × 10²⁰