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Atoms and Molecules
Physics IX : Complete NCERT Exercise Solutions
Access free NCERT Solutions for Class 9 Science Chapter 3: Atoms and Molecules. Understand laws of chemical combination, atomic & molecular mass, mole concept, and writing chemical formulas with step-by-step solved exercises and practice questions.
Question 3.1
A 0.24 g sample of a 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 & Explanation:
Given:
Mass of compound = 0.24 g
Mass of Boron = 0.096 g
Mass of Oxygen = 0.144 g
Percentage of Boron:
\( \text{% B} = \frac{\text{Mass of B}}{\text{Mass of compound}} \times 100 \)
\( = \frac{0.096}{0.24} \times 100 = 40\% \)
Percentage of Oxygen:
\( \text{% O} = \frac{0.144}{0.24} \times 100 = 60\% \)
Or by difference: 100% – 40% = 60%
Result: Boron = 40%, Oxygen = 60%
Mass of compound = 0.24 g
Mass of Boron = 0.096 g
Mass of Oxygen = 0.144 g
Percentage of Boron:
\( \text{% B} = \frac{\text{Mass of B}}{\text{Mass of compound}} \times 100 \)
\( = \frac{0.096}{0.24} \times 100 = 40\% \)
Percentage of Oxygen:
\( \text{% O} = \frac{0.144}{0.24} \times 100 = 60\% \)
Or by difference: 100% – 40% = 60%
Result: Boron = 40%, Oxygen = 60%
Question 3.2
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 combination will govern your answer?
Answer & Explanation:
Step 1: Determine the fixed combining ratio from first case:
Carbon + Oxygen → Carbon dioxide
3 g + 8 g → 11 g CO₂
Ratio C : O = 3 : 8 (by mass)
Step 2: In second case:
Carbon = 3 g, Oxygen = 50 g (excess)
Oxygen required for 3 g carbon = 8 g (from ratio)
Remaining oxygen = 50 – 8 = 42 g (unused)
Step 3: Mass of CO₂ produced:
Mass of CO₂ = Mass of C + Mass of O used
= 3 g + 8 g = 11 g
Step 4: Law governing:
Law of Constant Proportions (or Definite Proportions) – In a compound, elements always combine in fixed proportion by mass.
Carbon + Oxygen → Carbon dioxide
3 g + 8 g → 11 g CO₂
Ratio C : O = 3 : 8 (by mass)
Step 2: In second case:
Carbon = 3 g, Oxygen = 50 g (excess)
Oxygen required for 3 g carbon = 8 g (from ratio)
Remaining oxygen = 50 – 8 = 42 g (unused)
Step 3: Mass of CO₂ produced:
Mass of CO₂ = Mass of C + Mass of O used
= 3 g + 8 g = 11 g
Step 4: Law governing:
Law of Constant Proportions (or Definite Proportions) – In a compound, elements always combine in fixed proportion by mass.
Question 3.3
What are polyatomic ions? Give examples.
Answer & Explanation:
Polyatomic ions are charged particles (ions) that consist of two or more atoms covalently bonded together, carrying a net positive or negative charge.
Examples:
• Cations: Ammonium (NH₄⁺)
• Anions: Hydroxide (OH⁻), Nitrate (NO₃⁻), Carbonate (CO₃²⁻), Sulphate (SO₄²⁻), Phosphate (PO₄³⁻)
Examples:
• Cations: Ammonium (NH₄⁺)
• Anions: Hydroxide (OH⁻), Nitrate (NO₃⁻), Carbonate (CO₃²⁻), Sulphate (SO₄²⁻), Phosphate (PO₄³⁻)
Question 3.4
Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer & Explanation:
(a) Magnesium chloride: MgCl₂
(b) Calcium oxide: CaO
(c) Copper nitrate: Cu(NO₃)₂
(d) Aluminium chloride: AlCl₃
(e) Calcium carbonate: CaCO₃
Note: Copper has valency 2, nitrate is NO₃⁻.
(b) Calcium oxide: CaO
(c) Copper nitrate: Cu(NO₃)₂
(d) Aluminium chloride: AlCl₃
(e) Calcium carbonate: CaCO₃
Note: Copper has valency 2, nitrate is NO₃⁻.
Question 3.5
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer & Explanation:
(a) Quick lime (Calcium oxide, CaO): Calcium, Oxygen
(b) Hydrogen bromide (HBr): Hydrogen, Bromine
(c) Baking powder (Sodium bicarbonate, NaHCO₃): Sodium, Hydrogen, Carbon, Oxygen
(d) Potassium sulphate (K₂SO₄): Potassium, Sulphur, Oxygen
(b) Hydrogen bromide (HBr): Hydrogen, Bromine
(c) Baking powder (Sodium bicarbonate, NaHCO₃): Sodium, Hydrogen, Carbon, Oxygen
(d) Potassium sulphate (K₂SO₄): Potassium, Sulphur, Oxygen
Question 3.6
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
(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 & Explanation:
Atomic masses: C = 12 u, H = 1 u, S = 32 u, P = 31 u, Cl = 35.5 u, N = 14 u, O = 16 u
(a) C₂H₂: (2 × 12) + (2 × 1) = 24 + 2 = 26 g/mol
(b) S₈: 8 × 32 = 256 g/mol
(c) P₄: 4 × 31 = 124 g/mol
(d) HCl: 1 + 35.5 = 36.5 g/mol
(e) HNO₃: 1 + 14 + (3 × 16) = 1 + 14 + 48 = 63 g/mol
(a) C₂H₂: (2 × 12) + (2 × 1) = 24 + 2 = 26 g/mol
(b) S₈: 8 × 32 = 256 g/mol
(c) P₄: 4 × 31 = 124 g/mol
(d) HCl: 1 + 35.5 = 36.5 g/mol
(e) HNO₃: 1 + 14 + (3 × 16) = 1 + 14 + 48 = 63 g/mol
Question 3.7
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)?
(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 & Explanation:
(a) 1 mole of N atoms = Atomic mass in grams = 14 g
(b) 1 mole of Al atoms = 27 g
4 moles of Al atoms = 4 × 27 = 108 g
(c) Molar mass of Na₂SO₃:
= (2 × 23) + 32 + (3 × 16) = 46 + 32 + 48 = 126 g/mol
Mass of 10 moles = 10 × 126 = 1260 g
(b) 1 mole of Al atoms = 27 g
4 moles of Al atoms = 4 × 27 = 108 g
(c) Molar mass of Na₂SO₃:
= (2 × 23) + 32 + (3 × 16) = 46 + 32 + 48 = 126 g/mol
Mass of 10 moles = 10 × 126 = 1260 g
Question 3.8
Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Answer & Explanation:
Formula: \( \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \)
(a) O₂ molar mass = 32 g/mol
Moles of O₂ = \( \frac{12}{32} = 0.375 \) mol
(b) H₂O molar mass = 18 g/mol
Moles of H₂O = \( \frac{20}{18} \approx 1.11 \) mol
(c) CO₂ molar mass = 44 g/mol
Moles of CO₂ = \( \frac{22}{44} = 0.5 \) mol
(a) O₂ molar mass = 32 g/mol
Moles of O₂ = \( \frac{12}{32} = 0.375 \) mol
(b) H₂O molar mass = 18 g/mol
Moles of H₂O = \( \frac{20}{18} \approx 1.11 \) mol
(c) CO₂ molar mass = 44 g/mol
Moles of CO₂ = \( \frac{22}{44} = 0.5 \) mol
Question 3.9
What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer & Explanation:
Formula: Mass = Number of moles × Molar mass
(a) Molar mass of O atoms = 16 g/mol
Mass = 0.2 × 16 = 3.2 g
(b) Molar mass of H₂O = 18 g/mol
Mass = 0.5 × 18 = 9.0 g
(a) Molar mass of O atoms = 16 g/mol
Mass = 0.2 × 16 = 3.2 g
(b) Molar mass of H₂O = 18 g/mol
Mass = 0.5 × 18 = 9.0 g
Question 3.10
Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.
Answer & Explanation:
Step 1: Molar mass of S₈ = 8 × 32 = 256 g/mol
Step 2: Moles of S₈ = \( \frac{16}{256} = 0.0625 \) mol
Step 3: Number of molecules
= Moles × Avogadro's number
= 0.0625 × 6.022 × 10²³
= 3.76 × 10²² molecules
Step 2: Moles of S₈ = \( \frac{16}{256} = 0.0625 \) mol
Step 3: Number of molecules
= Moles × Avogadro's number
= 0.0625 × 6.022 × 10²³
= 3.76 × 10²² molecules
Question 3.11
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)
(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 & Explanation:
Step 1: Formula of aluminium oxide: Al₂O₃
Step 2: Molar mass of Al₂O₃:
= (2 × 27) + (3 × 16) = 54 + 48 = 102 g/mol
Step 3: Moles of Al₂O₃:
= \( \frac{0.051}{102} = 0.0005 \) mol = 5 × 10⁻⁴ mol
Step 4: Moles of Al³⁺ ions:
1 molecule Al₂O₃ contains 2 Al³⁺ ions
Moles of Al³⁺ = 2 × 5 × 10⁻⁴ = 10⁻³ mol
Step 5: Number of Al³⁺ ions:
= 10⁻³ × 6.022 × 10²³
= 6.022 × 10²⁰ ions
Step 2: Molar mass of Al₂O₃:
= (2 × 27) + (3 × 16) = 54 + 48 = 102 g/mol
Step 3: Moles of Al₂O₃:
= \( \frac{0.051}{102} = 0.0005 \) mol = 5 × 10⁻⁴ mol
Step 4: Moles of Al³⁺ ions:
1 molecule Al₂O₃ contains 2 Al³⁺ ions
Moles of Al³⁺ = 2 × 5 × 10⁻⁴ = 10⁻³ mol
Step 5: Number of Al³⁺ ions:
= 10⁻³ × 6.022 × 10²³
= 6.022 × 10²⁰ ions
📘 Exam Preparation Tip:
Practice numerical problems based on the mole concept, molecular mass, and formula unit mass. Memorize and apply the laws of chemical combination. Learn how to write chemical formulas and names of common compounds correctly.