AQA A-Level Chemistry AS Paper 1, June 2025: Question 21
1 mark · Medium difficulty · Multiple Choice
Calculate the volume of distilled water needed to dilute 25.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid to a concentration of 0.0400 mol dm⁻³.
Practise this questionQuestion
Question text
21 A student is asked to prepare a solution of hydrochloric acid with a concentration of
0.0400 mol dm–3
What volume of distilled water should the student add to
25.0 cm3 of 2.00 mol dm–3 hydrochloric acid to prepare this solution?
[1 mark]
A 25 cm3
B 50 cm3
C 1225 cm3
D 1250 cm3
Mark scheme
Show the mark scheme
21 C 1 (AO3) 1225 cm3
How to answer it
Dilution Calculations: Calculating Added Volume of Water
This question assesses your ability to perform accurate solution dilution calculations under timed conditions (AO3 practical and problem-solving skills):
- Applying the principle of conservation of moles during dilution: n = c × V .
- Using the dilution equation: c₁V₁ = c₂V₂ .
- Carefully distinguishing between total final solution volume and the volume of solvent added.
Solution Dilution Problem
Hydrochloric Acid Preparation
✅ Correct Answer
C — 1225 cm³
• Option C scores 1 mark (AO3).
• Requires calculating total volume (1250 cm³) and subtracting the initial acid volume (25.0 cm³).
💡 Key Knowledge
- When pure water is added to dilute an acid, the amount of solute (moles) remains constant:
moles before = moles after - Dilution relationship:
c₁ × V₁ = c₂ × V₂ - Relation between volumes:
V(water added) = V(total) - V(initial acid)
📐 Step-by-Step Calculation
Method 1: Using the Dilution Formula (c₁V₁ = c₂V₂)
- Identify the knowns:
Initial concentration, c₁ = 2.00 mol dm⁻³
Initial volume, V₁ = 25.0 cm³
Target concentration, c₂ = 0.0400 mol dm⁻³
Final total volume, V₂ = ? - Calculate total final volume (V₂):
V₂ = (c₁ × V₁) / c₂
V₂ = (2.00 × 25.0) / 0.0400 = 50.0 / 0.0400 = 1250 cm³ - Calculate volume of distilled water to add:
V(water) = V₂ - V₁
V(water) = 1250 cm³ - 25.0 cm³ = 1225 cm³
Method 2: Working via Moles
- Calculate moles of HCl in original solution:
Moles = concentration × volume (dm³) = 2.00 × (25.0 / 1000) = 0.0500 mol - Calculate total volume required for 0.0400 mol dm⁻³:
Total Volume (dm³) = moles / concentration = 0.0500 / 0.0400 = 1.25 dm³
Total Volume (cm³) = 1.25 × 1000 = 1250 cm³ - Subtract initial volume to find added water:
Volume of water added = 1250 cm³ - 25.0 cm³ = 1225 cm³
❌ Common Traps & Examiner Insight
- Selecting Option D (1250 cm³): This is the single most common distractor. Students correctly calculate the final volume of the solution ( 1250 cm³ ) but forget that the flask already contains 25.0 cm³ of acid. Always re-read: "What volume of distilled water should the student add..."
- Unit conversion confusion: In c₁V₁ = c₂V₂ , as long as concentrations match in units ( mol dm⁻³ ), the volumes can both be in cm³ . Converting unnecessarily to dm³ and back often introduces power-of-10 errors.
- Options A & B: Result from dividing concentrations upside down or misinterpreting the dilution factor (e.g. 2.00 / 0.0400 = 50 , and confusing the dilution factor with volume).
🧠 Exam Technique Checklist
- Highlight trigger words: Underline the word "add" in question stems involving dilution.
- Dilution factor shortcut: Notice that the solution is diluted from 2.00 to 0.0400 , which is a factor of 50 ( 2.00 / 0.0400 = 50 ). Therefore, final volume = 25.0 × 50 = 1250 cm³ . Water added = 1250 - 25.0 = 1225 cm³ .
- Sanity check: The added volume must be less than the total final volume.
Topics
Physical Chemistry · 3.1.2 Amount of Substance
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.