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1.5.9Principles of Chemistry

Calculate Concentrations of Solutions — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 21 of 25 Unit quiz

📚 Edexcel Linear - OL (Chemistry) Hub / Topical Revision
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Part 1 of 6Solutions and Concentration

A more concentrated solution holds more solute in the same volume.

Solute
  • Solute The substance that dissolves, which may be measured as a mass in grams or an amount in moles.
Solvent
  • Solvent The liquid, usually water, in which the solute dissolves.
Concentration
  • Concentration Concentration is the amount of solute in a stated volume of solution, most often per 1 dm³.
Units
  • Units The common units are g/dm³ for mass and mol/dm³ for amount of substance.

Part 2 of 6Equations to Use

Equations to Use
Moles formula
  • Moles formula Concentration (mol/dm³) = moles of solute ÷ volume of solution (dm³).
Rearrangements
  • Rearrangements Moles = concentration × volume, and volume = moles ÷ concentration.
Mass concentration
  • Mass concentration Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³).
Converting units
  • Converting units Multiply a concentration in mol/dm³ by Mr to obtain g/dm³, and divide g/dm³ by Mr to get mol/dm³.
Volume units
  • Volume units Divide a volume in cm³ by 1000 to get dm³ before using any concentration formula.
ConversionOperation
cm³ to dm³÷ 1000
dm³ to cm³× 1000
mol/dm³ to g/dm³× Mr

Part 3 of 6Working Through a Problem

Most errors come from skipping the unit conversion in the first step.

1
Convert every volume to dm³ by dividing the cm³ value by 1000.
2
Calculate moles of any substance with a known mass using mass ÷ Mr.
3
Use the balanced equation to link moles of one substance to another if a reaction is involved.
4
Apply concentration = moles ÷ volume, or the rearrangement needed.
5
State the answer with the correct unit and suitable significant figures.

Part 4 of 6Typical Uses

Typical Uses
Making a solution
  • Making a solution To make a given volume of a stated concentration, calculate the moles needed and then the mass using Mr.
Titration
  • Titration Concentrations of acids and alkalis are found by combining the moles ratio with measured volumes.
Comparing solutions
  • Comparing solutions Expressing concentrations in the same unit shows which solution has more solute per dm³.

Part 5 of 6Worked Example

Question

  • 4.0 g of sodium hydroxide is dissolved in water and made up to 250 cm³.
  • Calculate the concentration in g/dm³ and in mol/dm³.
  • (Ar: Na = 23, O = 16, H = 1)

  1. Volume = 250 ÷ 1000 = 0.250 dm³.
  2. Concentration in g/dm³ = 4.0 ÷ 0.250 = 16 g/dm³.
  3. Mr of NaOH = 23 + 16 + 1 = 40, so moles = 4.0 ÷ 40 = 0.10 mol.
  4. Concentration in mol/dm³ = 0.10 ÷ 0.250 = 0.40 mol/dm³ (check: 16 ÷ 40 = 0.40).

Answer 16 g/dm³, which is 0.40 mol/dm³

Part 6 of 6Exam Checklist & Tips

1.5.9 – Key Points for the Exam

  • Concentration Amount of solute dissolved per unit volume of solution.
  • Mole The amount of substance containing 6.02 × 10²³ particles, with mass equal to Mr in grams.
  • dm³ A cubic decimetre, equal to 1000 cm³ or one litre.
  • Solution A mixture of a solute dissolved in a solvent.

Common Mistakes

  • Using the volume in cm³ directly in the concentration formula.
  • Confusing g/dm³ with mol/dm³ and forgetting to divide by Mr.
  • Using the volume of solvent rather than the final volume of solution.
  • Writing answers without units, or with the wrong unit.

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