OCR A-Level Chemistry AS Breadth in chemistry (01), November 2021: Question 5

1 mark · Medium difficulty · Multiple Choice

Identify which gas sample has the greatest mass at RTP among four options.

Practise this question

Question

Multiple choice question 5 asks which gas sample has the greatest mass at RTP, with options A (50 cm3 of Ar(g)), B (100 cm3 of O2(g)), C (150 cm3 of N2(g)), and D (200 cm3 of Ne(g)). Below the options is an answer box and the mark allocation of [1].
Question text

5 Which gas sample has the greatest mass at RTP?

A 50 cm3 of Ar(g)

B 100 cm3 of O (g)

C 150 cm3 of N (g)

D 200 cm3 of Ne(g)

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme shows the correct answer for question 5 is C, worth 1 mark under assessment objective AO2.2.

5 C 1 AO2.2

How to answer it

Calculating the Greatest Mass of a Gas Sample at RTP

OCR AS Level Chemistry • Multiple Choice Question

What this question tests

This question assesses your ability to interconvert between gas volume, moles, and mass at Room Temperature and Pressure (RTP). Specifically, it tests AO2.2 (applying concepts of moles and molar volume to chemical data). To secure the mark, you must recognise that equal volumes of gases contain equal numbers of moles, but different masses depending on their relative molecular or atomic mass.

Question Analysis & Answer

Question 5 Multiple Choice

✅ Correct Answer: C

Option C ( 150 cm³ of N₂(g) ) yields the greatest mass when calculated step-by-step.

💡 Key Knowledge

  • At RTP, 1 mole of any gas occupies 24.0 dm³ (or 24,000 cm³ ).
  • Formula relating moles and volume: Moles = Volume / 24,000 cm³
  • Formula relating mass and moles: Mass = Moles × Molar Mass (Mr)

📐 Step-by-Step Calculation

  1. A: 50 cm³ of Ar (Ar = 39.9) → Moles = 50 / 24,000 = 0.00208 mol → Mass = 0.00208 × 39.9 = 0.0831 g
  2. B: 100 cm³ of O₂ (Mr = 32.0) → Moles = 100 / 24,000 = 0.00417 mol → Mass = 0.00417 × 32.0 = 0.133 g
  3. C: 150 cm³ of N₂ (Mr = 28.0) → Moles = 150 / 24,000 = 0.00625 mol → Mass = 0.00625 × 28.0 = 0.175 g (Highest Mass)
  4. D: 200 cm³ of Ne (Ne = 20.2) → Moles = 200 / 24,000 = 0.00833 mol → Mass = 0.00833 × 20.2 = 0.168 g

🧠 Exam Technique

Since the molar volume constant (24,000 cm³) applies to all options, you can shortcut the calculation by comparing the relative product of Volume × Mr for each gas directly, saving precious time in multiple-choice sections!

❌ Common Errors

  • Unit Mismatch: Forgetting to convert cm³ into dm³ (or failing to use 24,000 cm³ instead of 24 dm³).
  • Diatomic Gas Oversight: Using atomic mass for nitrogen ( N = 14.0 ) instead of molecular mass for nitrogen gas ( N₂ = 28.0 ). Oxygen ( O₂ = 32.0 ) is another frequent trap.
  • Intuition Trap: Students often incorrectly select option D simply because it has the largest volume ( 200 cm³ ), ignoring that Neon has a lower molar mass than Nitrogen.
Mark Scheme Allocation: 1 mark awarded for selecting C. (AO2.2)

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.