Concentration of Solutions
Concentration for IGCSE Chemistry 0620: mol/dm³ and g/dm³, n = cV, converting cm³ to dm³, and full titration calculations with mark breakdowns.
The IGCSE Chemistry Specialist Team · founded by Rig
Written to the Cambridge IGCSE Chemistry (0620) syllabus and mark-scheme conventions. Last updated 2026-06-11.
Titration questions on Paper 4 hand you 25.0 cm³ of alkali, a burette reading, and a 4-5 mark calculation that one missed ÷1000 wrecks from the first line. Concentration is Supplement content with exactly two units and one formula, yet it produces a disproportionate share of calculation casualties because the data arrives in cm³ and the formula wants dm³. The fix is procedural, and this page builds it from the mole foundations up.
Units and the core formula (Supplement)
Concentration measures how much solute is dissolved per unit volume of solution, in either g/dm³ or mol/dm³.
| Formula | Use |
|---|---|
| n = c × V | Moles from concentration and volume (V in dm³) |
| c = n ÷ V | Concentration from moles and volume |
| g/dm³ = mol/dm³ × Mr | Switching between the two units |
Dissolve 0.50 mol of NaOH to make 250 cm³ of solution: V = 0.250 dm³, c = 0.50 ÷ 0.250 = 2.0 mol/dm³. In g/dm³ that is 2.0 × 40 = 80 g/dm³. Run the conversion the other way by dividing by Mr.
The cm³ trap deserves its own paragraph. Laboratory glassware (pipettes at 25.0 cm³, burettes reading to 0.05 cm³) delivers cm³, and the formula requires dm³. Divide by 1000 first, write the converted value down: 25.0 cm³ = 0.0250 dm³. Treating it as a written step, not a mental one, is what stops the error.
Titration calculations (Supplement)
A titration finds an unknown concentration by reacting it with a solution of known concentration. The calculation is the standard three-step routine from reacting-mass calculations with solution conversions at both ends: moles of the known (n = cV), equation ratio, then c = n/V for the unknown.
Example: 25.0 cm³ of sodium hydroxide is neutralised by 20.0 cm³ of 0.100 mol/dm³ sulfuric acid. H2SO4 + 2NaOH → Na2SO4 + 2H2O. n(H2SO4) = 0.100 × 0.0200 = 0.00200 mol. Ratio 1:2, so n(NaOH) = 0.00400 mol. c(NaOH) = 0.00400 ÷ 0.0250 = 0.160 mol/dm³. The 1:2 ratio is the step students drop: sulfuric acid is diprotic, and the balanced equation, not intuition, supplies the 2. (The practical procedure, pipette, burette, indicator, is examined in topic 7 and Paper 6; the arithmetic lives here, and our mole calculations technique guide has a full drill ladder.)
Worked exam question
A student titrates 25.0 cm³ of dilute hydrochloric acid with 0.0500 mol/dm³ sodium carbonate solution. The mean titre is 22.50 cm³. The equation is Na2CO3 + 2HCl → 2NaCl + H2O + CO2. (a) Calculate the number of moles of sodium carbonate in the titre. (1) (b) Calculate the concentration of the hydrochloric acid in mol/dm³. (2) (c) Calculate this concentration in g/dm³. [Mr of HCl = 36.5] (1)
Model answer: (a) n = 0.0500 × 0.02250 = 1.125 × 10^−3 mol (1). (b) Ratio Na2CO3 : HCl = 1 : 2, so n(HCl) = 2.25 × 10^−3 mol (1); c = 2.25 × 10^−3 ÷ 0.0250 = 0.0900 mol/dm³ (1). (c) 0.0900 × 36.5 = 3.29 g/dm³ (1).
Mark-by-mark: (a) requires 22.50 cm³ converted to 0.02250 dm³ inside the substitution. (b) splits into the ratio-doubling mark and the division by the acid’s own volume. Using 22.50 cm³ again here is the classic crossed-volume error. (c) is a single ×Mr step; three significant figures is the expected rounding.
The mistakes that cost marks
- Volumes left in cm³. Every n = cV substitution needs dm³; one forgotten ÷1000 scales the answer by a thousand, an error magnitude that should fail your own sanity check.
- Crossed volumes: dividing the unknown’s moles by the other solution’s volume. Label each volume with its substance before calculating.
- Ratio amnesia with diprotic acids: H2SO4 and Na2CO3 questions almost always carry a 1:2 ratio, and skipping it halves or doubles the answer.
- Rounding too early. Keep 3-4 significant figures through intermediate lines; mark schemes specify a final-answer range that early rounding can exit.
How examiners want it phrased
| Typical student wording | Accepted mark-scheme wording |
|---|---|
| ”I used 25 in the formula" | "25.0 cm³ = 0.0250 dm³; n = c × V = 0.100 × 0.0250" |
| "The acid is 0.09 strong" | "Concentration of HCl = 0.0900 mol/dm³" |
| "Times two because of the equation" | "From the 1 : 2 mole ratio, n(HCl) = 2 × n(Na2CO3)" |
| "Convert to grams" | "Concentration in g/dm³ = 0.0900 mol/dm³ × 36.5 = 3.29 g/dm³” |
The Malaysia note
Titration is compulsory practical work in Malaysian international schools, but lab time and calculation time are budgeted separately, so students can run a burette competently yet freeze at the maths in the May/June Paper 4. The cure is pairing every practical with its calculation until the two feel like one task. Concentration work also returns in Paper 6 data questions, so the investment pays twice. Bring one failed titration calculation to a free trial lesson and a Chemistry specialist will locate the exact step that breaks.
Test yourself
Convert every volume to dm³ before you substitute. Then click each answer to check.
Q1 (2 marks). 4.0 g of sodium hydroxide is dissolved in water to make 500 cm³ of solution. Calculate the concentration in mol/dm³. [Mr of NaOH = 40]
Show answer
• n = 4.0 ÷ 40 = 0.10 mol [1] • c = 0.10 ÷ 0.500 = 0.20 mol/dm³ [1]
Q2 (3 marks). 25.0 cm³ of potassium hydroxide solution is exactly neutralised by 15.0 cm³ of 0.200 mol/dm³ nitric acid: HNO3 + KOH → KNO3 + H2O. Calculate the concentration of the potassium hydroxide in mol/dm³.
Show answer
• n(HNO3) = 0.200 × 0.0150 = 3.00 × 10^−3 mol [1] • Ratio 1 : 1, so n(KOH) = 3.00 × 10^−3 mol [1] • c(KOH) = 3.00 × 10^−3 ÷ 0.0250 = 0.120 mol/dm³ [1]
Q3 (2 marks). Calculate the concentration, in g/dm³, of a 0.0500 mol/dm³ solution of potassium chloride. [Ar: K 39, Cl 35.5]
Show answer
• Mr of KCl = 39 + 35.5 = 74.5 [1] • Concentration = 0.0500 × 74.5 = 3.73 g/dm³ [1]
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Frequently asked questions
What are the two concentration units in 0620?
g/dm³ (mass concentration) and mol/dm³ (molar concentration). They interconvert through the Mr: concentration in g/dm³ = concentration in mol/dm³ × Mr. Both are Supplement-level calculations.
What formula links moles, concentration and volume?
n = c × V, with volume in dm³. Rearranged: c = n ÷ V and V = n ÷ c. Since burettes and pipettes read in cm³, divide by 1000 before substituting, as the conversion is usually a marked step.
How are titration calculations structured?
Moles of the solution you know (n = cV), mole ratio from the balanced equation, then concentration of the unknown (c = n/V). It is the standard three-step mole method applied with solution conversions at both ends.