AQA A-Level Chemistry Paper 3, 2017: Question 30

1 mark · Medium difficulty · Multiple Choice

Calculate the volume of water needed to dilute 5.00 cm³ of 1.00 mol dm⁻³ ammonia solution to a concentration of 0.050 mol dm⁻³.

Practise this question

Question

Question 30 asks: 'A student is provided with 5.00 cm³ of 1.00 mol dm⁻³ ammonia solution. The student was asked to prepare an ammonia solution with a concentration of 0.050 mol dm⁻³. What volume of water should the student add?' Four multiple choice options are given: A 45.0 cm³, B 95.0 cm³, C 100 cm³, and D 995 cm³.
Question text

A student is provided with 5.00 cm3 of 1.00 mol dm−3 ammonia solution. The student

was asked to prepare an ammonia solution with a concentration of 0.050 mol dm−3

What volume of water should the student add?

[1 mark]

A 45.0 cm3

B 95.0 cm3

C 100 cm3

D 995 cm3

Mark scheme

Show the mark scheme Mark scheme table row showing question number 30 corresponds to correct answer B.

30 B

How to answer it

Solution Dilution & Volume Calculations

📋 What this question tests

This question assesses your ability to perform aqueous solution dilution calculations, specifically distinguishing between the total final volume of a prepared solution and the actual volume of solvent (water) that must be added to an existing concentrated stock solution.

Question 30 • Multiple Choice

Preparing a Dilute Ammonia Solution

Determining the volume of water required for a 20-fold dilution

✅ Correct Answer

B — 95.0 cm³

Mark Scheme: 1 mark for option B.

📐 Step-by-Step Calculation

Method 1: Dilution Factor Formula

  1. Calculate the dilution factor ( c₁ / c₂ ):
    1.00 / 0.050 = 20
  2. Calculate the total final volume ( V₂ ):
    V₂ = 5.00 cm³ × 20 = 100 cm³
  3. Calculate volume of water to add:
    V(water added) = V₂ - V₁
    V(water added) = 100 cm³ - 5.00 cm³ = 95.0 cm³

Method 2: Conservation of Amount (Moles)

  1. moles = c × V = 1.00 × (5.00 / 1000) = 5.00 × 10⁻³ mol
  2. V₂ = moles / c₂ = (5.00 × 10⁻³) / 0.050 = 0.100 dm³ = 100 cm³
  3. V(water) = 100 cm³ - 5.00 cm³ = 95.0 cm³

💡 Key Knowledge

  • Conservation of Solute: During dilution, only pure solvent is added. The amount of solute (moles) remains strictly constant:
    n₁ = n₂ ⇒ c₁ × V₁ = c₂ × V₂
  • Additive Volumes Assumption: In standard exam dilution scenarios, volumes are assumed to be additive:
    V_final = V_initial + V_water added
  • Units: While using c₁V₁ = c₂V₂ , as long as concentrations share identical units ( mol dm⁻³ ), volumes can remain in cm³ without converting to dm³ .

❌ Common Errors & Traps

  • Picking 100 cm³ (Option C): This is the most common pitfall! 100 cm³ is the total final volume of the solution, not the volume of water added.
  • Selecting 45.0 cm³ (Option A): Occurs if a student mistakenly calculates a 10-fold dilution instead of 20-fold ( 50 - 5 = 45 cm³ ).
  • Selecting 995 cm³ (Option D): Occurs from misreading 0.050 mol dm⁻³ as 0.0050 mol dm⁻³ , leading to a target total volume of 1000 cm³ ( 1000 - 5 = 995 cm³ ).

🧠 Examiner Insight & Technique

  • Read the Final Sentence Carefully: Multiple-choice questions frequently test precision in reading. Look for keywords such as "made up to" (total volume) versus "added to" (subtraction required).
  • Double-Check the Distractor: Examiners deliberately put the total volume ( 100 cm³ ) as an option because they know rushed candidates will stop calculating the moment their calculator screen displays 100 .

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.