AQA A-Level Chemistry AS Paper 1, June 2025: Question 22

Avogadro's Law & Atom Counting in Gases

🧪 WHAT THIS QUESTION TESTS

This multiple-choice question assesses your understanding of Avogadro’s Law, the distinction between gas molecules and constituent atoms (atomicity), and the ability to identify rapid deduction shortcuts rather than getting bogged down in time-consuming mass calculations.

Question 22

Multiple Choice Analysis

Identifying substances with equivalent total number of atoms

✅ Correct Answer: D (25.0 cm³ of oxygen)

According to Avogadro’s Law, equal volumes of any gases at identical temperature (T) and pressure (P) contain the exact same number of molecules.

25.0 cm³ of nitrogen gas (N₂) contains n molecules of N₂. Since each molecule is diatomic, it contains 2n atoms.
25.0 cm³ of oxygen gas (O₂) also contains n molecules of O₂. Since oxygen is also diatomic, it contains 2n atoms.

Therefore, 25.0 cm³ of O₂ contains precisely the same total number of atoms as 25.0 cm³ of N₂.

💡 Key Knowledge

Avogadro's Law: V ∝ n (at constant T and P). Equal volumes of all ideal gases contain equal numbers of moles of molecules.
Atomicity: Pay close attention to whether particles are mono-, di-, or polyatomic:
- Argon (Ar): 1 atom per particle
- Nitrogen (N₂) & Oxygen (O₂): 2 atoms per molecule
- Carbon dioxide (CO₂) & Sulfur dioxide (SO₂): 3 atoms per molecule

🧠 Exam Technique: Look for Shortcuts

Spot the distractor data: The mass 0.287 g is given to tempt you into calculating molar masses ( n = m / Mᵣ ) for all four options.
Check volume options first: Since both initial nitrogen and option D are diatomic gases occupying 25.0 cm³ at identical T and P, they must have the same number of atoms. You can answer this question in 10 seconds without picking up your calculator!

📐 Why the Other Options are Incorrect (Step-by-Step Proof)

Option A (25.0 cm³ of argon):
25.0 cm³ of Ar has the same number of molecules as 25.0 cm³ of N₂. However, argon is a noble gas (monatomic, Ar), so it contains only half as many atoms as diatomic N₂.
Option B (1.312 g of sulfur dioxide, SO₂):
Mᵣ(SO₂) = 32.1 + (2 × 16.0) = 64.1 g mol⁻¹
Moles of SO₂ = 1.312 / 64.1 = 0.0205 mol .
Each SO₂ molecule has 3 atoms → Total atoms = 3 × 0.0205 = 0.0615 mol of atoms.
Initial N₂: 0.287 / 28.0 = 0.01025 mol N₂ → 2 × 0.01025 = 0.0205 mol of atoms. (SO₂ has 3× too many atoms).
Option C (0.287 g of carbon dioxide, CO₂):
Mᵣ(CO₂) = 12.0 + (2 × 16.0) = 44.0 g mol⁻¹
Moles of CO₂ = 0.287 / 44.0 = 0.00652 mol .
Total atoms = 3 × 0.00652 = 0.0196 mol of atoms (≠ 0.0205 mol).

❌ Common Traps & Misconceptions

Confusing Atoms with Molecules: Choosing A because the volumes match (25.0 cm³), forgetting that Ar is monatomic while N₂ is diatomic.
Confusing Mass with Moles: Choosing C simply because the mass is identical ( 0.287 g ). Different substances have different molar masses, so equal masses rarely contain the same number of particles!
Wasting Time on Full Calculations: Performing long decimal arithmetic on Options B and C under timed exam conditions when Avogadro’s gas law directly gives the solution via Option D.

Mark Scheme Breakdown: [1 mark total] awarded for selecting D (AO2 application of chemical principles).

AQA A-Level Chemistry AS Paper 1, June 2025: Question 22

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Mark scheme

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