Concentration & dilution

3 practice questions

Amount and mass concentration

Learning goalUse concentration with the volume of the solution in the correct units.

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Concentration tells you how much solute is present in a given volume of solution. Use the final solution volume, not the volume of water added.

For concentration in mol dm⁻³, use c = n/V with V in dm³. Divide cm³ by 1000 to convert to dm³.

Mass concentration tells you grams of solute per dm³. Calculate mass divided by solution volume, or multiply molar concentration by molar mass: mass concentration = cM.

Write the concentration unit before choosing the calculation. mol dm⁻³ and g dm⁻³ describe different quantities.

c = n/V; n = cV; V = n/c

V solution volume, in dm³ when concentration is in mol dm⁻³; c concentration of the specified solute, in mol dm⁻³; n amount of the specified substance or particles, in mol.

Worked example

Sodium carbonate solution

5.30 g Na₂CO₃ is made up to 250 cm³ of solution. M = 106.0 g mol⁻¹.

Calculate the solution concentration in mol dm⁻³ and in g dm⁻³.

Worked answer

Quantity	Na₂CO₃
Mass / g	5.30
M / g mol⁻¹	106.0
Amount / mol	n = m/M 5.30/106.0 = 0.0500

V(dm³) = V(cm³)/1000
V = 250/1000 = 0.250 dm³; c = n/V
c = 0.0500/0.250 = 0.200 mol dm⁻³

Mass concentration = mass of solute ÷ solution volume
Mass concentration = 5.30/0.250

0.200 mol dm⁻³ and 21.2 g dm⁻³.

EXAM TRAPUse the final solution volume. Adding solute to 250 cm³ of water is not the same as making a solution up to 250 cm³.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 33

A solution contains 0.0850 mol Na₂CO₃ in a final volume of 250.0 cm³. M(Na₂CO₃) = 106.0 g mol⁻¹.

(a) Calculate its concentration in mol dm⁻³. (2 marks)

(b) Calculate its concentration in g dm⁻³. (2 marks)

(d) Explain why dissolving the solute in 250.0 cm³ of water would not necessarily produce the stated concentration. (1 mark)

Total: 7 marks

Show answer & marking points

(a) V = 0.2500 dm³ [1]; c = n/V
c = 0.0850/0.2500 = 0.340 mol dm⁻³. [1]

(b) Mass concentration = cM = 0.340 × 106.0 [1] = 36.0 g dm⁻³. [1]

(d) The required volume is the final solution volume, not the volume of solvent added. [1]

✓ DONE WHENUse concentration with the volume of the solution in the correct units.

Dilution and choosing a stock volume

Learning goalConserve solute amount during a dilution.

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Adding water spreads the same amount of solute through a larger volume. The concentration falls, but the solute's mole amount stays the same, assuming no reaction or loss.

That gives c₁V₁ = c₂V₂. The volumes must use matching units. If the stock volume is unknown, rearrange first to find the volume you need to measure.

Measure the required stock solution with suitable volumetric apparatus.
Transfer it completely to a volumetric flask.
Add deionised water to the calibration mark, using drops near the end.
Stopper and invert repeatedly to mix.

Make up to the required final volume. Do not simply add that volume of water to the stock solution.

Worked example

Dilute a stock solution

25.0 cm³ of 0.800 mol dm⁻³ NaCl is made up to 200.0 cm³.

Calculate the concentration of the diluted sodium chloride solution in mol dm⁻³.

Worked answer

n(NaCl) = cV
n(NaCl) = 0.800 × 0.0250 = 0.0200 mol

c₂ = n/V₂
c₂ = 0.0200/0.2000

0.100 mol dm⁻³.

To choose a stock volume: V₁ = c₂V₂/c₁

V₁ volume of stock solution used; use matching units for both volumes; V₂ final solution volume; use matching units for both volumes; c₁ initial stock concentration, in mol dm⁻³; c₂ final diluted concentration, in mol dm⁻³.

EXAM TRAPc₁V₁ = c₂V₂ applies to dilution of the same solute. A reaction between different substances needs a balanced equation and its mole ratio.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 34

25.00 cm³ of 0.8000 mol dm⁻³ NaOH is diluted to 250.0 cm³.

(a) Calculate the concentration of the diluted NaOH solution in mol dm⁻³. (2 marks)

(b) Calculate the volume, in cm³, of 0.500 mol dm⁻³ stock solution needed to prepare 100.0 cm³ of 0.0400 mol dm⁻³ solution. (2 marks)

Total: 6 marks

Show answer & marking points

(a) The solute amount is unchanged: c₂ = c₁V₁/V₂
c₂ = (0.8000 × 25.00)/250.0 [1] = 0.08000 mol dm⁻³. [1]

(b) V₁ = c₂V₂/c₁
V₁ = (0.0400 × 100.0)/0.500 [1] = 8.00 cm³. [1]

✓ DONE WHENConserve solute amount during a dilution.

Prepare a standard solution

Learning goalTransfer a known amount quantitatively into a measured solution volume.

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A standard solution has an accurately known concentration. To prepare one, both the mass of solute and the final solution volume must be accurate.

Calculate the required mass of a suitable pure, dry, stable solid. Weigh it accurately, using weighing by difference where appropriate.
Dissolve the solid fully in deionised water in a beaker, stirring with a glass rod.
Transfer the solution through a funnel into the volumetric flask.
Rinse the beaker, rod and funnel into the flask. This transfers solute that would otherwise be left behind.
Let a warm solution cool to room temperature. Add water nearly to the mark, then dropwise until the bottom of the meniscus is on the mark at eye level.
Stopper and invert repeatedly, then label the solution.

Quantitative transfer means transferring all the measured solute. A volumetric flask is calibrated to contain its stated volume at the calibration temperature.

Worked example

Plan a standard solution

Prepare 250.0 cm³ of 0.0500 mol dm⁻³ Na₂CO₃. M = 106.0 g mol⁻¹.

Calculate the mass of anhydrous sodium carbonate needed.

Worked answer

n = cV
n = 0.0500 × 0.2500 = 0.01250 mol

m = n × M
m = 0.01250 × 106.0

Weigh 1.325 g Na₂CO₃ (subject to the balance precision).

EXAM TRAPOvershooting the mark makes the concentration too low. Removing some mixed solution does not restore the required concentration; prepare a fresh solution.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 35

A student prepares 250.0 cm³ of 0.0600 mol dm⁻³ Na₂CO₃ from pure anhydrous solid. M = 106.0 g mol⁻¹.

(a) Calculate the mass needed. (2 marks)

(b) Describe an accurate preparation from the weighed solid to a uniform solution of the required volume. (6 marks)

Total: 10 marks

Show answer & marking points

(a) n = cV
n = 0.0600 × 0.2500 = 0.0150 mol [1]; m = n × M
m = 0.0150 × 106.0 = 1.59 g. [1]

(b) Dissolve the weighed solid fully in deionised water in a beaker. [1] Transfer to the correct volumetric flask. [1] Rinse the beaker, rod and funnel into the flask. [1] Add water near the mark, then dropwise. [1] Read the bottom of the meniscus at eye level on the mark. [1] Stopper and invert repeatedly to mix. [1]

(c) Let a warm solution cool to room temperature before making up to the mark. [1] If the mark is overshot, prepare a fresh solution; removing mixed solution does not restore its concentration. [1]

✓ DONE WHENTransfer a known amount quantitatively into a measured solution volume.

Concentration & dilution: OCR A A-level Chemistry