Concentration is the amount of solute in one dm³ of solution. Convert the volume to dm³ first, then divide the mass or moles by that volume.
This lesson is part of moles, formulas and equations. It uses mole ratios from balancing chemical equations.
What are the relationships?
| Quantity | Formula | Unit |
|---|---|---|
| Concentration by mass | mass of solute ÷ volume in dm³ | g dm⁻³ |
| Concentration by amount | moles of solute ÷ volume in dm³ | mol dm⁻³ |
| Moles from mass | mass ÷ relative formula mass | mol |
| Conversion | g dm⁻³ ÷ relative formula mass | mol dm⁻³ |
Volume in dm³ = volume in cm³ ÷ 1 000. Do this conversion on the first line of your working, every time.
Worked example 1: mass to concentration
4.0 g of sodium hydroxide (NaOH, relative formula mass 40) is dissolved to make 250 cm³ of solution.
- Volume = 250 ÷ 1 000 = 0.250 dm³.
- Moles = 4.0 ÷ 40 = 0.10 mol.
- Concentration = 0.10 ÷ 0.250 = 0.40 mol dm⁻³.
- In g dm⁻³: 4.0 ÷ 0.250 = 16 g dm⁻³. Check: 0.40 × 40 = 16. The two answers agree.
Worked example 2: dilution
25.0 cm³ of 2.0 mol dm⁻³ hydrochloric acid is diluted to 100 cm³. The amount of acid does not change, so M₁V₁ = M₂V₂.
2.0 × 25.0 = M₂ × 100, so M₂ = 50 ÷ 100 = 0.50 mol dm⁻³.
Worked example 3: a titration
25.0 cm³ of sodium hydroxide solution is neutralised by 20.0 cm³ of 0.10 mol dm⁻³ hydrochloric acid. The equation is NaOH + HCl → NaCl + H₂O, a 1 : 1 ratio.
- Moles of HCl = 0.10 × 20.0 ÷ 1 000 = 0.0020 mol.
- Moles of NaOH = 0.0020 mol, from the 1 : 1 ratio.
- Concentration of NaOH = 0.0020 ÷ 0.0250 = 0.080 mol dm⁻³.
Check the size: the base is less concentrated than the acid because more base volume was needed. That matches 0.080 being less than 0.10.
The slip that costs the most marks
The costliest slip is to leave the volume in cm³, which gives an answer 1 000 times too large. The second is to use the wrong ratio in a titration, such as H₂SO₄ with NaOH, which is 1 : 2.
Write the conversion first and check the ratio against the balanced equation before dividing. The concentration and dilution tutor lets you practise both on new numbers.
Check yourself
5.3 g of sodium carbonate (Na₂CO₃, relative formula mass 106) is dissolved to make 500 cm³ of solution. Find the concentration in mol dm⁻³ and in g dm⁻³.
Answer
Volume = 500 ÷ 1 000 = 0.500 dm³.
Moles = 5.3 ÷ 106 = 0.050 mol. Concentration = 0.050 ÷ 0.500 = 0.10 mol dm⁻³.
In g dm⁻³: 5.3 ÷ 0.500 = 10.6 g dm⁻³. Check: 0.10 × 106 = 10.6.
What to study next
Test the chain with the moles, formulas and equations practice set. Then see how these skills justify themselves in conservation reasoning across a reaction.
For a teacher to check your unit conversions as you work, see online one-to-one Chemistry tuition.