How to calculate number of moles from concentration and volume

Whenever a question gives you a solution, an acid, an alkali or a salt in water, and a volume, it is really handing you a number of moles. Turning “25 cm³ of 0.1 mol dm⁻³” into a mole value is one of the most-used skills in SPM Chemistry, and it comes down to a single formula used carefully. This guide sets out that formula, the one conversion that trips students up, and three worked examples, drawing on the mole concept chapter.

The one formula you need

n = M × V

• n is the number of moles of solute, in mol.
• M is the concentration (molarity), in mol dm⁻³.
• V is the volume of solution, in dm³.

That is it. Concentration is “moles per cubic decimetre”, so multiplying it by a volume in dm³ gives you moles. The whole difficulty is that volumes are almost always given in cm³, and you must convert first.

The conversion that costs marks

Volumes in SPM questions come in cubic centimetres, but the formula needs cubic decimetres. The rule is simple:

volume in dm³ = volume in cm³ ÷ 1000

So 25 cm³ = 0.025 dm³, and 250 cm³ = 0.25 dm³. Forgetting this single division is the single most common error in the whole topic, the answer comes out 1000 times too big. Convert to dm³ before you touch the formula, every time.

The method in three steps

1. Read off M, the concentration in mol dm⁻³.
2. Convert V to dm³, divide the given cm³ by 1000.
3. Multiply: n = M × V, and quote the answer in mol.

Worked example 1: a straightforward case

Find the number of moles of sodium hydroxide in 25.0 cm³ of 0.100 mol dm⁻³ NaOH.

• M = 0.100 mol dm⁻³.
• V = 25.0 ÷ 1000 = 0.0250 dm³.
• n = 0.100 × 0.0250 = 0.00250 mol.

That small number is correct, a quarter of a litre would hold 0.025 mol, so a tenth of that volume holds a tenth as much. The full walk-through of concentration units is on our molarity and concentration page.

Worked example 2: a larger volume

Find the moles of hydrochloric acid in 250 cm³ of 2.0 mol dm⁻³ HCl.

• M = 2.0 mol dm⁻³.
• V = 250 ÷ 1000 = 0.25 dm³.
• n = 2.0 × 0.25 = 0.50 mol.

Notice the sanity check: 2.0 mol dm⁻³ means 2 moles in a full 1 dm³, so a quarter of that volume holds a quarter of the moles, half a mole. A quick estimate like this catches a slipped decimal point.

Rearranging the formula

The same relationship answers two other question types, just rearranged:

• To find concentration: M = n ÷ V. If 0.0500 mol of solute is dissolved to make 0.500 dm³ of solution, M = 0.0500 ÷ 0.500 = 0.100 mol dm⁻³.
• To find volume: V = n ÷ M. If you need 0.0200 mol of a 0.400 mol dm⁻³ solution, V = 0.0200 ÷ 0.400 = 0.0500 dm³ = 50.0 cm³.

Practise spotting which quantity is missing, then rearrange to make it the subject.

Worked example 3: inside a titration

This formula is the engine of every titration calculation. Suppose 24.0 cm³ of 0.100 mol dm⁻³ NaOH exactly neutralises some hydrochloric acid.

• Moles of NaOH = 0.100 × (24.0 ÷ 1000) = 0.00240 mol.
• The equation NaOH + HCl → NaCl + H₂O is 1 : 1, so moles of HCl = 0.00240 mol.

From there you can find the acid’s concentration or volume. The complete titration method is on our titration calculation page.

Common mistakes to avoid

• Not converting cm³ to dm³. This is the big one, always divide by 1000 first.
• Confusing concentration with moles. Concentration is moles per dm³; it only becomes a number of moles once you multiply by the volume.
• Mismatching quantities. In a reaction, use the moles of the substance the question is about, and only convert between substances using the balanced equation.

Make the formula automatic

The whole skill is: identify M, convert V to dm³, and multiply. Do three of each type, find n, find M, find V, and the pattern locks in. If the cm³-to-dm³ conversion keeps slipping, that is a five-minute fix with a teacher; our online one-to-one lessons run in English from RM50 an hour with a paid one-hour trial, drilling these against real Paper 2 questions.