How to calculate molarity and do dilutions

Concentration questions reward students who are careful and punish those who rush their units. The chemistry itself is short: molarity is just moles per unit volume, and dilution is just spreading the same amount of solute through more water. Get those two ideas clear, keep your cm³ and dm³ apart, and this becomes a source of reliable marks. This guide takes you through molarity, converting to and from mass, and the dilution formula, with worked numbers throughout.

What molarity means

Molarity, also called molar concentration, is the number of moles of solute dissolved in one cubic decimetre of solution. Its unit is mol dm⁻³. The defining relationship is short:

molarity = moles of solute ÷ volume of solution in dm³, or M = n ÷ V

The single most important habit in this whole topic is that the volume must be in dm³, not cm³. Since 1 dm³ = 1000 cm³, you convert by dividing a cm³ volume by 1000. Forgetting this is the number-one cause of wrong answers, so make converting the volume your automatic first move. You can revise the core idea at molarity and concentration.

A worked example: from mass to molarity

4.0 g of sodium hydroxide is dissolved in water and made up to 500 cm³ of solution. Find the molarity.

• First find moles. Molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹, so moles = 4.0 ÷ 40 = 0.10 mol.
• Convert the volume: 500 cm³ = 500 ÷ 1000 = 0.500 dm³.
• Apply the formula: molarity = 0.10 ÷ 0.500 = 0.20 mol dm⁻³.

Two conversions and one division, nothing more. This is exactly the calculation you do when preparing a standard solution, and you can see the practical procedure at preparing a standard solution.

Converting between molarity and mass concentration

SPM sometimes gives concentration in g dm⁻³ instead of mol dm⁻³, and you may need to switch between them. The bridge is again the molar mass:

concentration in g dm⁻³ = molarity in mol dm⁻³ × molar mass

So a 0.20 mol dm⁻³ solution of NaOH has a concentration of 0.20 × 40 = 8.0 g dm⁻³. To go the other way, divide the g dm⁻³ value by the molar mass to recover the molarity. Keeping these two kinds of concentration distinct, one counts moles, one counts grams, stops a lot of confusion.

Dilution: the same solute, more water

When you dilute a solution you add water but do not add any more solute, so the number of moles of solute stays the same. Because moles are unchanged, the product of concentration and volume is the same before and after:

M₁V₁ = M₂V₂

Here M₁ and V₁ are the concentration and volume before dilution, and M₂ and V₂ are after. A convenient feature of this formula is that the volumes appear on both sides, so as long as you use the same unit for V₁ and V₂ they cancel, you can even leave both in cm³. Drill this at dilution of solutions.

Two dilution examples

Example one, find the new concentration. 25 cm³ of 2.0 mol dm⁻³ hydrochloric acid is diluted with water to 100 cm³. Then M₂ = M₁V₁ ÷ V₂ = (2.0 × 25) ÷ 100 = 0.50 mol dm⁻³. The concentration fell by a factor of four because the volume rose by a factor of four, a useful sanity check.

Example two, find the volume of stock needed. How much 1.0 mol dm⁻³ sodium chloride solution do you need to make 250 cm³ of 0.10 mol dm⁻³ solution? Rearrange to V₁ = M₂V₂ ÷ M₁ = (0.10 × 250) ÷ 1.0 = 25 cm³. So you measure 25 cm³ of the stock and add water up to 250 cm³. You can confirm answers like these with our molarity and dilution calculator.

The habits that keep you accurate

Three checks catch almost every error. Always convert volumes to dm³ before using M = n ÷ V, and only leave them in cm³ inside the dilution formula, where they cancel. Reach for the right relationship: use n = MV for a single solution and M₁V₁ = M₂V₂ only for a dilution. And do a quick reality check, a diluted solution must be weaker than the original, so if your new concentration comes out larger, you have inverted something. Master these and molarity and dilution become quietly reliable marks. If you would like a teacher to walk through standard-solution and dilution questions with you until the units are second nature, our online one-to-one lessons with our experienced SPM Chemistry teachers are built for that kind of practice, from RM50 an hour with a paid one-hour trial.