OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2024: Question 10

1 mark · Medium difficulty · Multiple Choice

Calculate the concentration of ethylammonium chloride formed when 1.35 g of ethylamine gas reacts with 20 cm3 of 2.0 mol dm-3 hydrochloric acid.

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Question

Multiple-choice question 10 asking to calculate the concentration of ethylammonium chloride in mol dm-3 formed from the reaction of 1.35 g of ethylamine gas (Mr = 45.0) with 20 cm3 of 2.0 mol dm-3 hydrochloric acid. The balanced equation given is CH3CH2NH2(g) + HCl(aq) -> CH3CH2NH3+(aq) + Cl-(aq). Four options are provided: A 0.03, B 0.67, C 1.50, and D 2.00, along with an answer box.
Question text

10 1.35 g of ethylamine gas, CH CH NH (M = 45.0), is reacted with 20 cm3 of 2.0 mol dm–3

32 2 r

hydrochloric acid forming a solution of ethylammonium chloride.

CH CH NH (g) + HCl(aq) CH CH NH +(aq) + Cl –(aq)

32 2 3 2 3

What is the concentration of ethylammonium chloride in mol dm–3?

A 0.03

B 0.67

C 1.50

D 2.00

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 10 is C, worth 1 mark, with 1.5(0) also allowed.

10 C 1 ALLOW 1.5(0)

How to answer it

Calculating Concentration of an Ammonium Salt Solution

What this question tests

  • Moles, mass, and molar mass calculations ( moles = mass ÷ Mᵣ ).
  • Using solution concentration and volume to calculate moles ( moles = concentration × volume in dm³ ).
  • Applying stoichiometric reacting ratios from a balanced chemical equation to identify the limiting reagent.
  • Calculating final concentration in mol dm⁻³ using the total solution volume.
Question 10 Multiple Choice Analysis

Solution & Step-by-Step Breakdown

✅ Correct Answer

C (1.50)

Mark Awarded: 1 / 1 (Allow 1.5)

💡 Key Knowledge

  • Amines act as weak bases, accepting protons from acids to form alkylammonium salts.
  • The balanced equation shows a 1 : 1 reacting ratio between ethylamine and hydrochloric acid.
  • Assuming the volume of added gas is negligible, the total volume of the resulting solution remains equal to the volume of the HCl solution used ( 20 cm³ ).

📐 Step-by-Step Calculation

  1. Calculate moles of ethylamine:
    Moles = 1.35 g ÷ 45.0 g mol⁻¹ = 0.030 mol
  2. Calculate moles of hydrochloric acid (HCl):
    Moles = 2.0 mol dm⁻³ × (20 ÷ 1000) dm³ = 0.040 mol
  3. Determine the limiting reagent and product moles:
    From the equation, CH₃CH₂NH₂(g) + HCl(aq) → CH₃CH₂NH₃⁺(aq) + Cl⁻(aq) , they react in a 1 : 1 ratio.
    Ethylamine ( 0.030 mol ) is the limiting reagent.
    Therefore, moles of CH₃CH₂NH₃⁺ formed = 0.030 mol .
  4. Calculate final concentration:
    Concentration = Moles ÷ Volume (dm³)
    Concentration = 0.030 mol ÷ 0.020 dm³ = 1.50 mol dm⁻³

❌ Common Errors & Traps

  • Distractor A (0.03): Forgetting to divide by the volume in dm³, leaving you with just the number of moles.
  • Distractor B (0.67): Confusing the stoichiometry or incorrectly mixing up reactant volumes and molar masses.
  • Distractor D (2.00): Assuming the concentration just stays equal to the original acid concentration without accounting for the added moles of amine.
  • Volume conversion trap: Forgetting to convert 20 cm³ into dm³ by dividing by 1000 .

🧠 Exam Technique & Examiner Insight

Examiners note that multi-step calculation questions in multiple-choice formats are designed to trap specific calculation errors. Always find the moles of both reactants first to quickly identify which one is in excess and which one limits the reaction product yield.

Topics

Module 2: Foundations in chemistry · Module 6: Organic chemistry and analysis · 6.2 Nitrogen compounds, polymers and synthesis · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.