AQA A-Level Chemistry Paper 3, 2021: Question 9

1 mark · Medium difficulty · Multiple Choice

Mole Fraction Calculation in a Binary Solution This multiple-choice question tests understanding of mole fraction calculations. It requires interpreting a binary mixture composed of a known mass of solute and solvent, converting masses to moles using relative molecular masses, and calculating the mole fraction of one component using the appropriate formula.

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Question

Question text

09 What is the mole fraction of 1.0 g of a compound of relative molecular mass 100.0

dissolved in 30.0 g of a solvent of relative molecular mass 50.0?

[1 mark]

A 6.0 × 10−3

B 1.6 × 10−2

C 1.7 × 10−2

D 3.0 × 10−2

Mark scheme

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09 B 1 1.6 × 10−2

How to answer it

Mole Fraction: Worked Example

Correct answer: B (1.6 × 10^–2)

Step-by-step:

Moles of solute = 1.0 g / 100.0 g mol⁻¹ = 0.010 mol
Moles of solvent = 30.0 g / 50.0 g mol⁻¹ = 0.60 mol
Total moles = 0.010 + 0.60 = 0.610 mol
Mole fraction of solute = 0.010 / 0.610 ≈ 1.6 × 10⁻²

Why the other options are wrong

A (6.0 × 10^–3): This underestimates the mole fraction — it may come from using total mass instead of total moles.

C (1.7 × 10^–2): A rounding error; very close, but B is the more accurate to 2 significant figures.

D (3.0 × 10^–2): This would be correct if the solvent moles were half the actual value — possibly a mistake in the molecular mass.

Tip: Always convert mass to moles before calculating mole fractions. Watch your sig figs and double-check that you're dividing by total moles, not grams.

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.