Topic 1 · page 5 of 10

Solutions and parts per million

Concentration is just amount per volume. Convert the volume to dm³ first, and a solution becomes another door into the mole.

11 min read1.51.4(iv)

Read first: Relative mass and the mole

Concentration in mol dm⁻³ 1.5

c=nVn=c×Vc=concentration / mol dm−3n=amount / molV=volume of solution / dm3 c = \frac{n}{V} \qquad n = c \times V \qquad \begin{aligned} c &= \text{concentration / mol dm}^{-3}\\ n &= \text{amount / mol}\\ V &= \text{volume of solution / dm}^3 \end{aligned}
A volume ladder: cubic metres, cubic decimetres and cubic centimetres. Moving from m³ to dm³ multiplies by 1000; from dm³ to cm³ multiplies by 1000. Moving the other way divides by 1000. 1 dm³ equals 1 litre equals 1000 cm³.
Each step down the ladder multiplies by 1000. Laboratory volumes are usually in cm³, so divide by 1000 to get dm³.

Worked exampleConcentration of a sodium hydroxide solution

2.00 g of NaOH is dissolved in water and made up to 250 cm³ of solution. Find its concentration.

  1. n = 2.00 ÷ 40.0 = 0.0500 mol
  2. V = 250 ÷ 1000 = 0.250 dm³
  3. c = 0.0500 ÷ 0.250 = 0.200 mol dm⁻³

Concentration in g dm⁻³ 1.5

Sometimes concentration is given as the mass of solute per dm³. Moving between the two units is one multiplication by the molar mass:

concentration / g dm−3=concentration / mol dm−3×M \text{concentration / g dm}^{-3} = \text{concentration / mol dm}^{-3} \times M

For the sodium hydroxide solution above: 0.200 × 40.0 = 8.00 g dm⁻³. You can check this directly: 2.00 g ÷ 0.250 dm³ = 8.00 g dm⁻³.

Worked exampleMaking up a hydrated salt solution

What mass of copper(II) sulfate crystals, CuSO₄·5H₂O, is needed to make 250 cm³ of 0.100 mol dm⁻³ solution?

  1. n = c × V = 0.100 × 0.250 = 0.0250 mol
  2. M(CuSO₄·5H₂O) = 249.6 g mol⁻¹
  3. m = 0.0250 × 249.6 = 6.24 g

Parts per million 1.4(iv)

For very small amounts, percentages become awkward strings of zeros. Parts per million (ppm) counts how many parts of a substance there are in every million parts of the whole mixture.

ppm=amount of the substanceamount of the whole mixture×106 \text{ppm} = \frac{\text{amount of the substance}}{\text{amount of the whole mixture}} \times 10^{6}

The "amount" must be measured in the same way on the top and the bottom:

Gases in the atmosphere

Gas Approximate level in dry air As a percentage
nitrogen 780 000 ppm 78%
oxygen 210 000 ppm 21%
carbon dioxide about 420 ppm, and rising 0.042%
methane about 1.9 ppm 0.00019%

Converting is a shift of four powers of ten, because 1% = 10 000 ppm:

ppm=%×104 \text{ppm} = \% \times 10^{4}

Worked exampleCarbon dioxide in a room

How much CO₂ is there in 1.00 dm³ of air containing 420 ppm CO₂?

  1. 1.00 dm³ = 1000 cm³
  2. Volume of CO₂ = 1000 × 420 ÷ 10⁶ = 0.420 cm³

Check yourself

  1. How many moles of HCl are in 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid?
  2. What volume of 0.200 mol dm⁻³ sodium hydroxide solution contains 0.0100 mol of NaOH? Give your answer in cm³.
  3. A solution of sodium chloride is 0.500 mol dm⁻³. What is its concentration in g dm⁻³?
  4. A solution contains 9.80 g dm⁻³ of sulfuric acid, H₂SO₄. What is its concentration in mol dm⁻³?
  5. A 500 g sample of water contains 0.0035 g of lead. What is the lead content in ppm?
  6. The safe limit for carbon monoxide in a room is 50 ppm. What volume of CO, in cm³, is that in 2.0 m³ of air? Express 50 ppm as a percentage.

Answers cover these until you have tried the questions

  1. n = 0.100 × (25.0 ÷ 1000) = 2.50 × 10⁻³ mol
  2. V = n ÷ c = 0.0100 ÷ 0.200 = 0.0500 dm³ = 50.0 cm³
  3. 0.500 × 58.5 = 29.3 g dm⁻³
  4. M(H₂SO₄) = 98.1 g mol⁻¹. 9.80 ÷ 98.1 = 0.0999 mol dm⁻³ (about 0.100)
  5. 0.0035 ÷ 500 × 10⁶ = 7.0 ppm (2 significant figures, as in the data)
  6. 2.0 m³ = 2.0 × 10⁶ cm³. Volume of CO = 2.0 × 10⁶ × 50 ÷ 10⁶ = 100 cm³ (1.0 × 10² cm³). As a percentage: 50 ÷ 10⁴ = 0.0050%.