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Bridge the Gap · Module 4

Solutions & Concentration

Titrations make a return at the start of Year 12. Combined science students especially should familiarise themselves with the basics.

Joe White AQA & OCR examiner 15–25 minutes · self-marking
250 cm³ c = n ÷ V

c = n ÷ V

Concentration is moles per volume: c = n ÷ V, with c in mol/dm3 and V in dm3 — that last part is where the marks go missing. Lab volumes arrive in cm3, so the ÷1000 conversion happens inside the calculation, before anything else. You'll also meet concentration in g/dm3: multiply mol/dm3 by Mr to convert.

Concentration and titrations are a Chemistry-only GCSE topic — the groundwork is in C3 — Concentration & titrations.

c (mol/dm3) = n (mol)V (dm3)   ·   g/dm3 = mol/dm3 × Mr

Worked example — 0.025 mol dissolved in 50 cm3
  1. Volume to dm3 first:

    V = 50 ÷ 1000 = 0.050 dm3

  2. Then the equation:

    c = 0.025 ÷ 0.050 = 0.50 mol/dm3

Concentrated vs dilute

Same solute, more water. Concentration measures how crowded the solute is — keep the amount fixed and add water, and the same particles spread through more solution, so c falls and the colour fades. (A bigger volume does not mean more concentrated — that's the trap.) Drag the slider to add water and watch c = n ÷ V.

c = 0.50 mol ÷ 0.25 dm3 = 2.0 mol/dm3

Concentrated — the solute (0.50 mol) never changes; only the water does.

diluteconcentrated

Here the solute stays put and the water changes. For the other lever — fixed volume, more solute — the particle model in C3 — Concentration & titrations is worth a look.

Practice

🧪 Try it — concentration
Q1

0.10 mol dissolved in 200 cm3 — concentration in mol/dm3?

mol/dm3
Worked steps
  1. V = 200 ÷ 1000 = 0.200 dm3
  2. c = 0.10 ÷ 0.200 = 0.50 mol/dm3
Q2

Moles in 25.0 cm3 of 0.100 mol/dm3 solution?

mol
Worked steps
  1. Rearrange: n = c × V
  2. V = 25.0 ÷ 1000 = 0.0250 dm3
  3. n = 0.100 × 0.0250 = 0.0025 mol (= 2.5 × 10−3)
Q3

What volume (dm3) of 0.60 mol/dm3 solution contains 0.030 mol?

dm3
Worked steps
  1. V = n ÷ c
  2. 0.030 ÷ 0.60 = 0.050 dm3 (= 50 cm3)
Q4

4.0 g of NaOH dissolved to 500 cm3 — concentration in mol/dm3? Mr 40.0. Two steps.

mol/dm3
Worked steps
  1. n = m ÷ Mr = 4.0 ÷ 40.0 = 0.10 mol
  2. V = 500 ÷ 1000 = 0.500 dm3
  3. c = 0.10 ÷ 0.500 = 0.20 mol/dm3
Q5

A 0.20 mol/dm3 solution of NaOH — concentration in g/dm3? Mr 40.0.

g/dm3
Worked steps
  1. g/dm3 = mol/dm3 × Mr
  2. 0.20 × 40.0 = 8.0 g/dm3
Q6

Which is the same concentration as 0.50 mol/dm3? Tick (✓) one box.

How did you do?

Work through the questions above and your score appears here.

This module grows over the summer — more titration-style practice is on its way. Next: percentage yield and error, the habits A-Level practical work expects.

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