Concentration & dilution
Concentration & dilution
Solution concentration and mass concentration
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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 concentration in mol dm⁻³ by Mr to get mass concentration in g dm⁻³.
Write the concentration unit before choosing the calculation. mol dm⁻³ and g dm⁻³ describe different quantities.
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.
| Quantity | Unit |
|---|---|
| n | mol |
| c | mol dm−3 |
| V in n = cV | dm3 |
V solution volume, in dm³ when concentration is in mol dm⁻³.
Amount in an aliquot
Calculate the number of moles in 25.0 cm3 of 0.200 mol dm−3 NaCl.
Worked answer
V = 25.0/1000 = 0.0250 dm3
n = 0.200 × 0.0250
Find the required volume
What volume of 0.150 mol dm−3 KNO3 contains 0.0120 mol?
Worked answer
V = 0.0120/0.150 = 0.0800 dm3
Mr relative molecular or formula mass; no unit; c concentration of the specified solute, in mol dm⁻³.
Mass and molar concentration
5.85 g NaCl is dissolved and made up to 500.0 cm3. Mr(NaCl) = 58.5.
Calculate the solution concentration in mol dm⁻³ and in g dm⁻³.
Worked answer
n = 5.85/58.5 = 0.100 mol; V = 0.5000 dm3
c = 0.100/0.5000 = 0.200 mol dm−3
Mass concentration = 5.85/0.5000 = 11.7 g dm−3
Mass needed for a solution
Prepare 250.0 cm3 of 0.0800 mol dm−3 KCl. Mr(KCl) = 74.6.
Calculate the mass of KCl needed.
Worked answer
n = 0.0800 × 0.2500 = 0.0200 mol
m = 0.0200 × 74.6 = 1.492 g
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
(a) Calculate the number of moles in 32.50 cm3 of 0.1800 mol dm−3 KOH. (2 marks)
(b) Calculate the concentration, in mol dm⁻³, of a solution containing 0.01440 mol in 200.0 cm3. (2 marks)
(c) Calculate the volume, in cm3, of 0.1250 mol dm−3 solution that contains 0.00750 mol solute. (2 marks)
Total: 6 marks
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(a) V(dm³) = V(cm³)/1000
V = 32.50/1000 = 0.03250 dm3 [1]; n = cV = 0.1800 × 0.03250 = 0.005850 mol [1].
(b) V = 0.2000 dm3 [1]; c = n/V = 0.01440/0.2000 = 0.07200 mol dm−3 [1].
(c) V = n/c = 0.00750/0.1250 = 0.0600 dm3 [1] = 60.0 cm3 [1].
(a) A solution is made by dissolving 2.70 g K2SO4 and making the solution up to 500.0 cm3. Calculate its concentration in mol dm−3. Mr(K2SO4) = 174.3. (3 marks)
(b) Calculate the mass concentration of this solution in g dm−3. (1 mark)
(c) Calculate the mass of CaCl2 needed to prepare 250.0 cm3 of 0.0200 mol dm−3 solution. Mr(CaCl2) = 111.0. (2 marks)
Total: 6 marks
Show answer & marking points
(a) n(K₂SO₄) = m/Mᵣ
n(K2SO4) = 2.70/174.3 = 0.01549 mol [1]; V = 0.5000 dm3 [1]; c = n/V
c = 0.01549/0.5000 = 0.03098 mol dm−3, or 0.0310 mol dm−3 [1].
Mass concentration = mass of solute ÷ solution volume
(b) Mass concentration = 2.70/0.5000 = 5.40 g dm−3 [1].
(c) n = cV = 0.0200 × 0.2500 = 0.00500 mol [1]; m = nMr = 0.00500 × 111.0 = 0.555 g [1].
Dilute a solution accurately
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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.
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.
Find the diluted concentration
Dilute 20.0 cm3 of 0.500 mol dm−3 solution to 100.0 cm3.
Calculate the concentration of the diluted solution in mol dm⁻³.
Worked answer
c2 = (0.500 × 20.0)/100.0
Measure the stock volume
Prepare 250.0 cm3 of 0.0400 mol dm−3 solution from 0.200 mol dm−3 stock.
Calculate the volume of stock solution needed, in cm³, and state the final volume after dilution.
Worked answer
V1 = (0.0400 × 250.0)/0.200
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
(a) 15.0 cm3 of 0.8000 mol dm−3 NaCl is diluted to a final volume of 250.0 cm3. Calculate the final concentration in mol dm⁻³. (3 marks)
(b) Calculate the volume, in cm³, of 0.500 mol dm−3 stock solution needed to prepare 100.0 cm3 of 0.0650 mol dm−3 solution. (2 marks)
(c) Explain why the stock volume from part (b) must not be mixed with 100.0 cm3 of water. (1 mark)
Total: 6 marks
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(a) n before = 0.8000 × 0.0150 = 0.0120 mol [1]; final V = 0.2500 dm3 [1]; c = n/V
c = 0.0120/0.2500 = 0.0480 mol dm−3 [1].
(b) c1V1 = c2V2, so V₁ = c₂V₂/c₁
V1 = (0.0650 × 100.0)/0.500 [1] = 13.0 cm3 [1].
(c) 100.0 cm3 is the final solution volume; adding 100.0 cm3 of water would make the final volume greater than 100.0 cm3 and the solution too dilute [1].
Prepare a standard solution
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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.
Sodium carbonate standard
Prepare 250.0 cm3 of 0.0500 mol dm−3 anhydrous Na2CO3. Mr = 106.0.
Calculate the mass of anhydrous sodium carbonate needed.
Worked answer
n = 0.0500 × 0.2500 = 0.0125 mol
m = 0.0125 × 106.0
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Prepare 250.0 cm³ of 0.0600 mol dm⁻³ Na₂CO₃ from a suitable anhydrous solid. Mᵣ(Na₂CO₃) = 106.0.
- Calculate the required mass. (2 marks)
- Describe how to prepare the solution accurately, including weighing and transfer. (6 marks)
Total: 8 marks
Show answer & marking points
(a) n = cV
n = 0.0600 × 0.2500 = 0.0150 mol [1]; m = n × Mᵣ
m = 0.0150 × 106.0 = 1.59 g. [1]
(b) Weigh accurately, for example by difference, and record the actual mass. [1] Dissolve fully in deionised water in a beaker. [1] Transfer through a funnel to the volumetric flask and add the beaker/rod/funnel washings. [1] Let a warm solution cool to the flask’s calibration temperature. [1] Make up to the mark dropwise, with the bottom of the meniscus at eye level. [1] Stopper and invert repeatedly to mix. [1]