Concentration

The concentration of a solution tells you how much solute is dissolved in each dm³ (litre) of it. In g/dm³ it is the mass in each dm³; in mol/dm³ it is the number of moles. Multiply by the Mᵣ to go from mol/dm³ to g/dm³. Adding water dilutes a solution: the moles stay the same in a bigger volume.

Key facts

  • 1 dm³ = 1 litre = 1000 cm³: divide a volume in cm³ by 1000 to get dm³.
  • concentration (g/dm³) = mass (g) ÷ volume (dm³).
  • concentration (mol/dm³) = moles ÷ volume (dm³), so moles = concentration × volume.
  • To change mol/dm³ into g/dm³, multiply by the Mᵣ.
  • Diluting with water keeps the moles of solute the same, so the concentration falls as the volume grows.

Worked examples

Example 1

20 g of sodium chloride is dissolved in water to make 2 dm³ of solution. What is the concentration in g/dm³?

  1. Concentration = mass of solute ÷ volume of solution in dm³. concentration = 20 ÷ 2 = 10 g/dm³

Answer: 10 g/dm³

Example 2

5 g of copper sulfate is dissolved to make 250 cm³ of solution. Find the concentration in g/dm³.

  1. There are 1000 cm³ in 1 dm³. 250 ÷ 1000 = 0.25 dm³
  2. Divide the mass by the volume in dm³. 5 ÷ 0.25 = 20 g/dm³

Answer: 20 g/dm³

Example 3

What mass of sugar is dissolved in 500 cm³ of a sugar solution with a concentration of 40 g/dm³?

  1. Change the volume to dm³. 500 ÷ 1000 = 0.5 dm³
  2. Multiply the concentration by the volume. 40 × 0.5 = 20 g

Answer: 20 g

Example 4

Which is more concentrated: 6 g of salt in 200 cm³ of solution, or 10 g of salt in 500 cm³ of solution?

  1. 6 g in 0.2 dm³. 6 ÷ 0.2 = 30 g/dm³
  2. 10 g in 0.5 dm³. 10 ÷ 0.5 = 20 g/dm³
  3. More mass in each dm³ means more concentrated, so the first solution is.

Answer: 6 g in 200 cm³ (30 g/dm³, compared with 20 g/dm³)

Example 5

4 g of sodium hydroxide is dissolved to make 250 cm³ of solution. Find its concentration in g/dm³ and in mol/dm³. (Mᵣ of NaOH = 40)

  1. Change the volume to dm³. 250 ÷ 1000 = 0.25 dm³
  2. Concentration in g/dm³. 4 ÷ 0.25 = 16 g/dm³
  3. Moles of NaOH. 4 ÷ 40 = 0.1 mol
  4. Concentration in mol/dm³. 0.1 ÷ 0.25 = 0.4 mol/dm³

Answer: 16 g/dm³; 0.4 mol/dm³

Example 6

0.5 mol of hydrogen chloride is dissolved to make 2 dm³ of hydrochloric acid. What is its concentration in mol/dm³, and how many moles of HCl are in 100 cm³ of it?

  1. Concentration = moles ÷ volume in dm³. 0.5 ÷ 2 = 0.25 mol/dm³
  2. Moles = concentration × volume in dm³. 0.25 × 0.1 = 0.025 mol

Answer: 0.25 mol/dm³; 0.025 mol

Example 7

How many moles of solute are in 50 cm³ of a 2 mol/dm³ solution?

  1. Volume: 50 ÷ 1,000 = 0.05 dm³
  2. n = c × V = 2 × 0.05 = 0.1 mol

Answer: 0.1 mol

Example 8

10 cm³ of 1 mol/dm³ acid is diluted with water to 200 cm³. What is the new concentration in mol/dm³?

  1. Moles of acid = 1 × 10 ÷ 1,000 = 0.01 mol
  2. New concentration = 0.01 ÷ 0.2 = 0.05 mol/dm³

Answer: 0.05 mol/dm³

Example 9

A solution of chem NaOH has a concentration of 0.1 mol/dm³. What is its concentration in g/dm³? (Aᵣ: Na = 23, O = 16, H = 1)

  1. Mᵣ = 23 + 16 + 1 = 40
  2. 0.1 × 40 = 4 g/dm³

Answer: 4 g/dm³

Example 10

A solution of chem NaOH contains 80 g/dm³. What is its concentration in mol/dm³? (Aᵣ: Na = 23, O = 16, H = 1)

  1. Mᵣ = 23 + 16 + 1 = 40
  2. 80 ÷ 40 = 2 mol/dm³

Answer: 2 mol/dm³