Concentration of Solutions Exam Questions
Practice IGCSE Chemistry exam questions on concentration of solutions. Covers calculating concentration in g/dm3 and mol/dm3, dilution calculations, and converting between mass and volume with mark schemes and examiner notes.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
These questions cover all aspects of concentration of solutions. Show all working clearly.
Question 1 (2 marks, Core)
A student dissolves 10.0 g of sodium chloride in water to make 500 cm3 of solution.
Calculate the concentration of the solution in g/dm3.
Mark scheme
- Convert volume: 500 cm3 = 0.5 dm3 [1]
- Concentration = mass / volume = 10.0 / 0.5 = 20.0 g/dm3 [1]
Examiner note: Always convert cm3 to dm3 first (divide by 1000). A very common error is dividing 10.0 by 500 to get 0.02 g/cm3, which is a different unit. The question asks for g/dm3.
Question 2 (3 marks, Supplement)
Calculate the concentration in mol/dm3 of a solution made by dissolving 5.85 g of sodium chloride (NaCl) in water to make 250 cm3 of solution.
(Ar: Na = 23, Cl = 35.5)
Mark scheme
- Mr of NaCl = 23 + 35.5 = 58.5 [1]
- Moles of NaCl = 5.85 / 58.5 = 0.1 mol [1]
- Volume = 250/1000 = 0.25 dm3; concentration = 0.1 / 0.25 = 0.4 mol/dm3 [1]
Examiner note: For mol/dm3, you must convert mass to moles first, then divide by volume in dm3. The sequence is: mass → moles (÷ Mr) → concentration (÷ volume in dm3).
Question 3 (2 marks, Core)
A solution of hydrochloric acid has a concentration of 36.5 g/dm3.
(a) Calculate the mass of HCl in 2.0 dm3 of this solution. (1)
(b) Calculate the mass of HCl in 100 cm3 of this solution. (1)
Mark scheme
(a) mass = concentration x volume = 36.5 x 2.0 = 73.0 g [1]
(b) Volume = 100/1000 = 0.1 dm3; mass = 36.5 x 0.1 = 3.65 g [1]
Examiner note: Rearrange concentration = mass / volume to get mass = concentration x volume. Always check units — volume must be in dm3 when using g/dm3 concentration.
Question 4 (3 marks, Supplement)
A student needs to make 250 cm3 of 0.2 mol/dm3 sodium hydroxide solution.
Calculate the mass of sodium hydroxide (NaOH) needed. (Ar: H = 1, O = 16, Na = 23)
Mark scheme
- Volume = 250/1000 = 0.25 dm3 [1]
- Moles needed = concentration x volume = 0.2 x 0.25 = 0.05 mol [1]
- Mr of NaOH = 40; mass = 0.05 x 40 = 2.0 g [1]
Examiner note: Work backwards: concentration and volume → moles → mass. This is a common practical question. The student would dissolve 2.0 g of NaOH in enough water to make the total volume up to 250 cm3.
Question 5 (3 marks, Supplement)
Convert a concentration of 4.0 g/dm3 for sodium hydroxide (NaOH) into mol/dm3.
(Ar: H = 1, O = 16, Na = 23)
Mark scheme
- Mr of NaOH = 23 + 16 + 1 = 40 [1]
- Concentration in mol/dm3 = concentration in g/dm3 / Mr [1]
- = 4.0 / 40 = 0.1 mol/dm3 [1]
Examiner note: To convert g/dm3 to mol/dm3, divide by Mr. To go the other way (mol/dm3 to g/dm3), multiply by Mr. This conversion is frequently tested.
Question 6 (4 marks, Supplement)
25.0 cm3 of 0.10 mol/dm3 sodium hydroxide solution exactly neutralises 20.0 cm3 of hydrochloric acid.
NaOH + HCl → NaCl + H2O
(a) Calculate the number of moles of NaOH used. (1)
(b) Deduce the number of moles of HCl that reacted. (1)
(c) Calculate the concentration of the hydrochloric acid in mol/dm3. (2)
Mark scheme
(a) Moles of NaOH = (25.0/1000) x 0.10 = 0.0025 mol [1]
(b) From the equation, mole ratio NaOH : HCl = 1 : 1, so moles of HCl = 0.0025 mol [1]
(c) Volume of HCl = 20.0/1000 = 0.020 dm3 [1] Concentration = 0.0025 / 0.020 = 0.125 mol/dm3 [1]
Examiner note: This is a titration calculation — one of the most important types in IGCSE. The method is: moles of known → mole ratio → moles of unknown → concentration of unknown. Always convert volumes to dm3.
Question 7 (3 marks, Core)
A student has 1.0 dm3 of copper sulfate solution with a concentration of 80 g/dm3. She takes 250 cm3 of this solution.
(a) Calculate the mass of copper sulfate in the 250 cm3 sample. (1)
(b) She adds water to the 250 cm3 sample to make the total volume up to 1.0 dm3. Calculate the new concentration in g/dm3. (2)
Mark scheme
(a) Volume = 250/1000 = 0.25 dm3; mass = 80 x 0.25 = 20 g [1]
(b) The mass of solute stays the same = 20 g [1] New concentration = 20 / 1.0 = 20 g/dm3 [1]
Examiner note: Dilution does not change the mass of solute — it only increases the volume. The new concentration = original mass of solute / new volume. Here the volume increased by a factor of 4, so the concentration decreased by a factor of 4 (from 80 to 20 g/dm3).
What to revise if you scored below 5
If unit conversions caused errors, memorise: 1 dm3 = 1000 cm3. If concentration formulae were confused, use the triangle: C = n/V (or C = mass/V for g/dm3). For the titration question (Question 6), practise the four-step method. Revisit the concentration of solutions notes.
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