OCR A-Level Chemistry AS Breadth in chemistry (01), June 2023: Question 10

1 mark · Medium difficulty · Multiple Choice

Calculate the concentration of dilute hydrochloric acid prepared by diluting 15.00 cm3 of 18.0 mol dm-3 concentrated hydrochloric acid to 250 cm3.

Practise this question

Question

Multiple choice question 10 asking to find the concentration in mol dm-3 of dilute hydrochloric acid formed by diluting 15.00 cm3 of 18.0 mol dm-3 concentrated hydrochloric acid to 250 cm3. Four options are given: A 0.0675, B 0.270, C 0.300, and D 1.08. An empty answer box and a mark allocation of [1] appear at the bottom right.
Question text

10 15.00 cm3 of 18.0 mol dm–3 concentrated hydrochloric acid is diluted with water to prepare

250 cm3 of dilute hydrochloric acid.

What is the concentration, in mol dm–3, of the dilute hydrochloric acid?

A 0.0675

B 0.270

C 0.300

D 1.08

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 10 is option D, worth 1 mark.

10 D 1 AO2.4

How to answer it

Calculating Concentration After Dilution

What this question tests

This question assesses your ability to apply the dilution formula ( c₁V₁ = c₂V₂ ) or calculate moles of solute before and after dilution. It tests unit conversions between cm³ and dm³, and manipulation of concentration equations under Assessment Objective 2 (Applying chemical knowledge).

Question 10

Solution & Breakdown

✅ Correct Answer

D (1.08)

Mark Scheme: Option D [1 mark] (AO2.4)

💡 Key Knowledge

  • Dilution formula: c₁V₁ = c₂V₂ (where moles of solute remain constant).
  • Conversion factor: 1 dm³ = 1000 cm³ .
  • Concentration units: mol dm⁻³ .

📐 Step-by-Step Calculation

  1. Find the moles of HCl in the initial concentrated solution:
    Moles = concentration × volume (in dm³)
    Moles = 18.0 × (15.00 / 1000) = 0.270 mol
  2. Use the new total volume to find the new concentration:
    New Concentration ( c₂ ) = Moles / New Volume (in dm³)
    c₂ = 0.270 / (250 / 1000) = 0.270 / 0.250 = 1.08 mol dm⁻³
  3. Alternative using the dilution formula directly:
    c₂ = (c₁ × V₁) / V₂
    c₂ = (18.0 × 15.00) / 250 = 270 / 250 = 1.08 mol dm⁻³
    *(Note: Because both volumes are in cm³, the conversion factors cancel out, so you don't strictly need to convert them to dm³ if both are in the same units!)*

🧠 Exam Technique

Look at the distractor values! Many common student calculation errors are intentionally placed as multiple-choice options (A, B, and C). Always double-check your unit conversions and make sure your final answer makes logical sense (diluting an 18.0 mol dm⁻³ solution should result in a much smaller concentration, but not impossibly tiny).

❌ Common Errors & Distractors

  • Distractor B (0.270): This is just the number of moles calculated in step 1 ( 18.0 × 0.015 ). Students often stop halfway through the calculation.
  • Unit Mix-ups: Forgetting to divide cm³ by 1000 when calculating moles separately, leading to magnitude errors.

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), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.