Principles of Chemistry

8 marks · Normal

This question is about using moles and gas molar volumes in calculations. Ammonium chloride decomposes on heating according to the following equation: NH4Cl(s) NH3(g) + HCl(g)

(a)

Define the mole in terms of the number of particles.

[2 marks]

(b)

Recall the volume occupied by 1 mole of any gas at rtp in: (i) dm³ (ii) cm³

[2 marks]

(c)

Calculate the total volume of gas (in cm³) produced at rtp when 5.35 g of ammonium chloride decomposes completely. (Ar: N = 14, H = 1, Cl = 35.5; molar volume = 24,000 cm³ mol⁻¹)

[4 marks]

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(a) One mole is the amount of substance that contains exactly 6 × 10²³ particles (Avogadro's number). · must show Avogadro constant or equivalent definition

(b) i) 24 dm³ (ii) 24,000 cm³ · must show both units correct

Common mistakes: Forgetting to multiply by number of moles of gas produced

Show mark scheme and worked solution

Marking points

B1 (a) · 2 marks · Defines the mole using Avogadro's number or equivalent.

B2 (b) · 1 mark · States 24 dm³ as the volume for 1 mol of gas at rtp.

B3 (b) · 1 mark · States 24,000 cm³ as the volume for 1 mol of gas at rtp.

M1 (c) · 1 mark · Calculates Mr of NH4Cl: (14 + 4×1 + 35.5 = 53.5). (follow-through)

M2 (c) · 1 mark · Calculates moles: 5.35 / 53.5 = 0.100 mol. (follow-through)

M3 (c) · 1 mark · Recognises 0.100 mol NH4Cl gives 0.100 mol NH3 + 0.100 mol HCl (=0.200 mol gas). (follow-through)

A1 (c) · 1 mark · Calculates total gas volume: 0.200 × 24,000 = 4,800 cm³, with correct units. (follow-through)

Worked solution

a. One mole is the amount of substance containing 6.0 × 10²³ particles (atoms, molecules, ions); this is Avogadro's number.

b. i) 1 mole of any gas at rtp = 24 dm³; ii) 24,000 cm³.

c. Step 1: Mr of NH4Cl = 14 + (4×1) + 35.5 = 53.5 Moles = 5.35 / 53.5 = 0.100 mol Step 2: Each decomposes to 1 mol NH3 and 1 mol HCl → so 0.100 mol produces 0.100 mol NH3 + 0.100 mol HCl = 0.200 mol total gas.

Step 3: Total gas volume = 0.200 × 24,000 = 4,800 cm³.