OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2021: Question 6

1 mark · Medium difficulty · Multiple Choice

Calculate the volume of water needed to dilute a solution of magnesium iodide to a given iodide ion concentration.

Practise this question

Question

Multiple choice question 6 asking for the volume of water to add when diluting 100.0 cm3 of 0.400 mol dm-3 magnesium iodide to an iodide concentration of 0.250 mol dm-3. Four options are given: A 60.0, B 160.0, C 220.0, D 320.0, along with an answer box.
Question text

6 A student is supplied with 100.0 cm3 of a solution of 0.400 mol dm–3 magnesium iodide, MgI .

A student plans to dilute this solution so that the iodide concentration is 0.250 mol dm–3.

What volume of water, in cm3, does the student need to add?

A 60.0

B 160.0

C 220.0

D 320.0

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme table shows question number 6 with the correct answer as C, worth 1 mark under assessment objective AO2.6.

6 C 1 AO2.6

How to answer it

Solution Dilution and Ion Concentration Calculations

What this question tests

This question assesses your understanding of solution chemistry, specifically dealing with ionic stoichiometry during dilutions. It tests whether you can relate the concentration of an ionic salt to the concentration of its constituent ions (stoichiometric ratios) and distinguish between total solution volume and the volume of water added.

Question 6 Analysis

Multiple-Choice Mark Scheme: Option C

✅ Correct Answer

C (220.0 cm³) is the correct final answer.

💡 Key Knowledge

  • Formula for magnesium iodide: MgI₂ .
  • Dissociation equation: MgI₂ (aq) → Mg²⁺ (aq) + 2I⁻ (aq) .
  • The concentration of iodide ions (I⁻) is twice the concentration of the MgI₂ solution.
  • Dilution formula: C₁V₁ = C₂V₂ .
  • Crucial distinction: The question asks for the volume of water added, not the final total volume.

🧠 Exam Technique

  • Always write out ionic dissociation equations when a question specifies the concentration of a specific ion rather than the whole compound.
  • Underline key constraint words in the stem, such as "volume of water, in cm³, does the student need to add?" to avoid stopping halfway through the calculation.

❌ Common Errors

  • The 1:1 Trap (Option A or B): Forgetting that MgI₂ releases 2 moles of I⁻ per 1 mole of salt, leading to incorrect initial ion concentrations.
  • The Final Volume Trap (Option B - 160.0 cm³): Calculating the final total volume ( 320.0 cm³ ) and subtracting the initial volume incorrectly, or mistakenly selecting the final volume itself without subtracting the starting 100.0 cm³ .

📐 Step-by-Step Calculation

  1. Find the initial concentration of iodide ions (I⁻):
    Since MgI₂ → Mg²⁺ + 2I⁻ , every 1 mol of MgI₂ provides 2 mol of I⁻ .
    Initial [MgI₂] = 0.400 mol dm⁻³
    Initial [I⁻] = 0.400 × 2 = 0.800 mol dm⁻³
  2. Apply the dilution formula ( C₁V₁ = C₂V₂ ) to find the final total volume ( V₂ ):
    C₁ = 0.800 mol dm⁻³
    V₁ = 100.0 cm³
    C₂ = 0.250 mol dm⁻³ (target iodide concentration)
    V₂ = (C₁ × V₁) / C₂
    V₂ = (0.800 × 100.0) / 0.250 = 320.0 cm³
  3. Calculate the volume of water to add:
    Volume of water added = Final total volume − Initial volume
    Volume of water = 320.0 cm³ - 100.0 cm³ = 220.0 cm³
Examiner Note: This question effectively separates students who blindly apply C₁V₁ = C₂V₂ from those who properly analyze ion stoichiometry and carefully read the final requirement ("volume of water to add" vs. "final volume"). Top-tier candidates easily navigated the 1:2 mole ratio for iodide ions.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

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