A student prepares a solution of copper sulfate (CuSO₄) by dissolving 0.20 mol of copper sulfate in water to make 250 cm³ of solution. Explain how the concentration of this solution would change if the student had instead dissolved the same 0.20 mol of copper sulfate in water to make 500 cm³ of solution.

AQA GCSE Chemistry (8462) — 4.3.4 Using concentrations of solutions in mol/dm³ (Chem only) · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

The concentration would be lower because the same 0.20 mol of CuSO₄ is now dissolved in a larger volume. Concentration is moles ÷ volume (dm³). The original solution: 0.20 mol ÷ 0.250 dm³ = 0.80 mol dm⁻³. If the volume is doubled to 0.500 dm³, the new concentration is 0.20 mol ÷ 0.500 dm³ = 0.40 mol dm⁻³, which is half the original value.

Examiner tips

  • Show the formula C = n/V and calculate both concentrations; mention the volume doubles
  • Explain that the number of moles is unchanged, so the concentration must decrease

Common mistakes

  • Using cm³ instead of dm³ in the calculation
  • Failing to state that the concentration halves when the volume doubles

Mark scheme (4 marks)

  1. The concentration would decrease / be lower / be smaller
  2. Because the volume of solution has increased (doubled) / the same number of moles is dissolved in a larger volume
  3. Concentration = moles ÷ volume (in dm³), so the original concentration is 0.80 mol/dm³
  4. The new concentration is 0.40 mol/dm³ (0.20 mol ÷ 0.500 dm³)

Key terms in this question

concentration

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