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How do we measure and calculate the concentration of a solution?

Concentration of solutions in g/dm3 and mol/dm3; converting between them; using concentration in calculations; and titration ideas.

A focused answer to AQA GCSE Chemistry 4.3.4, covering concentration in g/dm3 and mol/dm3, converting between mass, volume and concentration, and how concentration is used in titration calculations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
What concentration means
Concentration in g/dm3
Concentration in mol/dm3 (Higher)
Titrations (Higher)
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What this dot point is asking

AQA wants you to define concentration, calculate it in g/dm $^3$ and in mol/dm $^3$ , convert between the two, and use concentration in calculations including titrations. Concentration links the mole work of 4.3 to practical acid-alkali chemistry: it tells you how many moles of a substance are packed into each cubic decimetre of solution, which is exactly what you need before you can apply a mole ratio. The mol/dm $^3$ work and titration calculations are Higher tier.

What concentration means

The unit of volume matters more than anything else in these calculations. A burette and pipette measure in cm $^3$ , but the concentration formula needs dm $^3$ , so the very first habit to build is dividing any volume in cm $^3$ by 1000 before substituting. A $25.0$ cm $^3$ pipette delivers $0.025$ dm $^3$ .

Concentration in g/dm3

For example, $20$ g of salt in $0.5$ dm $^3$ gives $20 / 0.5 = 40$ g/dm $^3$ . This is the only concentration foundation candidates need; it requires no moles, just mass divided by volume.

Concentration in mol/dm3 (Higher)

\text{concentration (mol/dm}^3) = \frac{\text{moles of solute}}{\text{volume (dm}^3)}

To convert from g/dm $^3$ to mol/dm $^3$ , divide by the relative formula mass; to go the other way, multiply by $M_r$ . This is because g/dm $^3$ and mol/dm $^3$ describe the same solution measured in two different "currencies", mass and moles, linked by the relative formula mass exactly as in the moles equation. For example, a solution of $40$ g/dm $^3$ of $NaOH$ ( $M_r = 40$ ) is $40 / 40 = 1$ mol/dm $^3$ .

Titrations (Higher)

A titration is used to find the volume of acid and alkali that react exactly. An indicator (or a pH probe) marks the end point. From the volumes and one known concentration, you can find the unknown concentration:

Calculate the moles of the substance you know (concentration $\times$ volume in dm $^3$ ).
Use the mole ratio from the equation.
Divide by the volume of the unknown to find its concentration.

Finding an unknown concentration by titration

$25.0$ cm $^3$ of potassium hydroxide solution is neutralised by $30.0$ cm $^3$ of $0.20$ mol/dm $^3$ sulfuric acid: $H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O$ . Find the concentration of the $KOH$ .

step 1 Convert the known volume to dm3

Volume of acid $= 30.0 / 1000 = 0.030$ dm $^3$ .

step 2 Find the moles of the known substance

Moles of $H_2SO_4 = 0.20 \times 0.030 = 0.0060$ mol.

step 3 Apply the mole ratio

The equation shows $1$ mol of $H_2SO_4$ reacts with $2$ mol of $KOH$ , so moles of $KOH = 2 \times 0.0060 = 0.012$ mol.

step 4 Divide by the volume of the unknown

Volume of $KOH = 25.0 / 1000 = 0.025$ dm $^3$ . Concentration $= 0.012 / 0.025 = 0.48$ mol/dm $^3$ .

Try this

Q1. Calculate the concentration in g/dm $^3$ of $5$ g of solute in $250$ cm $^3$ of solution. [2 marks]

Cue. $250$ cm $^3 = 0.25$ dm $^3$ ; concentration $= 5 / 0.25 = 20$ g/dm $^3$ .

Q2. State how to convert a concentration from g/dm $^3$ to mol/dm $^3$ . [1 mark]

Cue. Divide by the relative formula mass.

Q3. A $0.5$ mol/dm $^3$ solution of $HCl$ ( $M_r = 36.5$ ) is made. Calculate its concentration in g/dm $^3$ . [2 marks]

Cue. $0.5 \times 36.5 = 18.25$ g/dm $^3$ .

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20204 marksA solution is made by dissolving

14.6

g of sodium chloride in water and making the total volume up to

500

^3

. Calculate the concentration of the solution in mol/dm

^3

. Relative atomic masses: Na = 23, Cl = 35.5.

Show worked answer →

A 4-mark Higher tier calculation: marks for $M_r$ , moles, volume conversion, and the final concentration.

The $M_r$ of $NaCl = 23 + 35.5 = 58.5$ (1 mark). Moles $= 14.6 / 58.5 = 0.25$ mol (1 mark). Volume $= 500 / 1000 = 0.5$ dm $^3$ (1 mark). Concentration $= 0.25 / 0.5 = 0.5$ mol/dm $^3$ (1 mark).

The most common error is forgetting to convert cm $^3$ to dm $^3$ , giving an answer 1000 times too small. Markers accept the working in any logical order as long as the volume is in dm $^3$ .

AQA 20225 marksIn a titration,

25.0

^3

of sodium hydroxide solution was exactly neutralised by

20.0

^3

0.10

mol/dm

^3

hydrochloric acid. The equation is

HCl + NaOH \rightarrow NaCl + H_2O

. Calculate the concentration of the sodium hydroxide solution in mol/dm

^3

Show worked answer →

A 5-mark Higher titration question, with marks for volume conversions, moles of acid, the mole ratio, and the final concentration.

Moles of $HCl = \text{concentration} \times \text{volume} = 0.10 \times (20.0 / 1000) = 0.0020$ mol (2 marks). The equation shows a $1:1$ ratio, so moles of $NaOH = 0.0020$ mol (1 mark). Volume of $NaOH = 25.0 / 1000 = 0.025$ dm $^3$ (1 mark). Concentration of $NaOH = 0.0020 / 0.025 = 0.08$ mol/dm $^3$ (1 mark).

Markers reward correct unit handling throughout. A frequent slip is dividing the acid moles by the wrong volume.

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Sources & how we know this

AQA GCSE Chemistry (8462) specification — AQA (2016)