Vidaara.orgClass 9 · Chemistry
CodeVID-C9-03-T3-01
Assignment — Mole Concept & Molar Mass
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Use N_A = 6.022×10²³ and atomic masses H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, Ca = 40.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
One mole of any substance contains:
- A.6.022×10²² particles
- B.6.022×10²³ particles
- C.12 particles
- D.100 particles
2.
The molecular mass of H2O is:
- A.16 u
- B.17 u
- C.18 u
- D.20 u
3.
The molar mass of NaCl (Na = 23, Cl = 35.5) is:
- A.48.5 g
- B.58.5 g
- C.68.5 g
- D.35.5 g
4.
The number of moles in 64 g of oxygen gas O2 (molar mass 32 g) is:
- A.1
- B.2
- C.4
- D.0.5
5.
The formula unit mass concept is used for:
- A.gases only
- B.ionic compounds
- C.single atoms
- D.noble gases
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define molar mass and Avogadro number.
7.
Calculate the number of moles in 11 g of carbon dioxide CO₂ (molar mass 44 g).
8.
Find the molecular mass of ammonia NH₃ (N = 14, H = 1).
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Calculate the number of molecules in 4.4 g of carbon dioxide CO₂ (molar mass 44 g).
10.
Find the mass of 3.011×10²³ atoms of sodium (Na = 23).
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
A sample contains 5.85 g of sodium chloride NaCl (molar mass 58.5 g). Calculate (a) the number of moles, (b) the number of formula units, and (c) the total number of ions present.
Answer Key
Section A — Multiple Choice Questions
- (B) 6.022×10²³ particles
- (C) 18 u
- (B) 58.5 g
- (B) 2
- (B) ionic compounds
Section B — Short Answer (2 marks)
- Molar mass is the mass of one mole of a substance in grams (numerically equal to its atomic/molecular mass in u). The Avogadro number is the number of particles in one mole, 6.022×10²³.
- n = mass ÷ molar mass = 11 ÷ 44 = 0.25 mole.
- Molecular mass = 14 + 3(1) = 17 u.
Section C — Short Answer (3 marks)
- Moles = 4.4 ÷ 44 = 0.1 mole; molecules = 0.1 × 6.022×10²³ = 6.022×10²².
- Moles = 3.011×10²³ ÷ 6.022×10²³ = 0.5 mole; mass = 0.5 × 23 = 11.5 g.
Section D — Long Answer (5 marks)
- (a) Moles = 5.85 ÷ 58.5 = 0.1 mole. (b) Formula units = 0.1 × 6.022×10²³ = 6.022×10²². (c) Each NaCl gives one Na⁺ and one Cl⁻ (2 ions), so total ions = 2 × 6.022×10²² = 1.2044×10²³ ions.
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