RBSE Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Rajasthan Board RBSE Solutions for Class 9 Science Chapter 3 Atoms and Molecules Textbook Exercise Questions and Answers.

RBSE Class 9 Science Solutions Chapter 3 Atoms and Molecules

RBSE Class 9 Science Chapter 3 Atoms and Molecules InText Questions and Answers

Page No. 32

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, “Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction” is the result of the law of conservation of mass.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
The postulate of Dalton, “The relative number and kinds of atoms are fixed in a given compound” can explain the law of definite proportions.

Page No. 35

Question 1.
Define 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’.

Question 2.
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 No. 39

Question 1.
Write down the formulae of:
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide
Answer:

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) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate

Question 3.
What is meant by the term chemical formula?
Answer:
The chemical formula of a compound is a symbolic representation of its composition.

Question 4.
How many atoms are present in a (i) H₂S molecule and (ii) \(\mathrm{PO}_{4}^{3-}\) ion?
Answer:
(i) In a H₂S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a \(\mathrm{PO}_{4}^{3-}\) ion, five atoms are present; one of phosphorus and four of oxygen.

Page No. 40

Question 1.
Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₂, 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 0
= 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 = 30u
Molecular mass of C₂H₄ = 2 × Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1 = 28u
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 + 3 × Atomic mass of H + Atomic mass of O + Atomic mass of H = 12 + 3 × 1 + 8 + 1 = 24u

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
= 78 + 12 + 48 = 138 u

Page No. 42

Question 1.
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 1 mole of carbon atoms = 12 g
Then, mass of 6.022 × 10²³ number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon = 12 ÷ (6.022 × 10²³)
= 1.9926 × 10^–23 g

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 = 6.022 × 10²³ number of atoms
Thus, 100 g of Na contains = \(\frac{6.022 \times 10^{23}}{23}\) × 100 number of atoms
= 2.6182 × 10²⁴ number of atoms
Again, atomic mass of Fe = 56 u (Given)
Then, gram atomic mass of Fe = 56 g
Now, 56 g of Fe contains = 6.022 × 10²³ number of atoms
Thus, 100 g of Fe contains = \(\frac{6.022 \times 10^{23}}{56}\) × 100 number of atoms
= 1.0753 × 10²⁴ number of atoms
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

RBSE Class 9 Science Chapter 3 Atoms and Molecules Textbook Questions and Answers

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:
Total mass of compound = 0.24 g (Given)
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 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 / 0.24 × 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:
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 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 : Nitrate (NO₃^–), hydroxide ion (OH^–).

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:

Question 5.
Give the names of the elements present in the following compounds :
(a) Quicklime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer:
(a) Calcium and oxygen (CaO)
(b) Hydrogen and bromine (HBr)
(c) Sodium, hydrogen, carbon, and oxygen (NaHCO₃)
(d) Potassium, sulphur, and oxygen (K₂SO₄)

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 = 26g
(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 l 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 = 1.111 mole
(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 × 16 g = 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 = \(\frac{6.022 \times 10^{23}}{256}\) × 16 molecules
= 3.76375 × 10²² molecules.

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:
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 = \(\frac{6.022 \times 10^{23}}{102}\) × 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 0.051 g of aluminium oxide (Al₂O₃) = 2 × 3.011 × 10²⁰ = 6.022 × 10²⁰ molecules.