A cup of coffee and an espresso both contain caffeine — but one hits harder. That difference is concentration. Every time a nurse measures an IV drip rate, a pool technician tests chlorine levels, or a chemist prepares a reagent, they're working with this concept. It is the single most used calculation in Year 11 and 12 Chemistry.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
If you dissolve one teaspoon of sugar in a small cup of water versus the same teaspoon in a swimming pool, the amount of sugar is the same — but something is very different. What is different, and how would you describe that difference using a mathematical expression? What two quantities would you need to measure?
Type your initial thoughts below:
Record your ideas in your workbook.
📚 Core Content
Wrong: The limiting reagent is the one present in the smallest mass.
Right: The limiting reagent is the reactant that runs out first based on mole ratios, not mass.
A solution is a homogeneous mixture of a solute (the substance being dissolved) in a solvent (the dissolving medium, usually water). Concentration tells you how much solute is packed into a given volume of solution.
In chemistry, concentration is almost always expressed in moles per litre (mol L⁻¹) — also called molarity. This unit connects the macroscopic world (litres of solution you can measure in a lab) to the microscopic world (moles of particles) via a single formula.
The volume in c = n ÷ V must be in litres. Laboratory volumes are almost always given in millilitres. This conversion must become automatic.
Sometimes concentration is given or required in grams per litre (g L⁻¹). To convert between the two, use molar mass as the bridge.
🧮 Worked Examples
🧪 Activities
1 Calculate the concentration of a solution made by dissolving 0.400 mol of glucose in 2.00 L of solution.
2 How many moles of KNO₃ are in 400 mL of a 0.250 mol L⁻¹ solution?
3 What volume of 0.500 mol L⁻¹ sulfuric acid contains 0.0750 mol of H₂SO₄?
4 A solution of copper sulfate (CuSO₄) has a concentration of 16.0 g L⁻¹. Express this in mol L⁻¹. (Cu = 63.546, S = 32.06, O = 15.999)
Type your working for all four problems:
Answer in your workbook.
| Solution | Moles of solute (mol) | Volume (mL) | Volume (L) | Concentration (mol L⁻¹) |
|---|---|---|---|---|
| NaOH | 0.500 | 250 | ? | ? |
| HCl | ? | 100 | ? | 2.00 |
| CaCl₂ | 0.0400 | ? | ? | 0.160 |
| H₂SO₄ | 1.50 | 600 | ? | ? |
| KMnO₄ | ? | 50.0 | ? | 0.0200 |
Show all working below:
Complete in your workbook — show all working.
At the start of this lesson, you described the difference between dissolving sugar in a small cup versus a swimming pool, and thought about what quantities you'd need to measure.
The answer: concentration = amount of solute ÷ volume of solution, or c = n ÷ V (in mol L⁻¹). The amount of sugar (solute) stays the same, but the volume of the solution is much larger in the pool — so the concentration is far lower. You need to measure both the number of moles of solute and the volume in litres.
Reflect: how close was your initial idea to the formula c = n ÷ V?
Write a reflection in your workbook.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. A nurse prepares an intravenous saline solution by dissolving 9.00 g of NaCl in water to produce 1.00 L of solution. Calculate the concentration of the solution in (a) mol L⁻¹ and (b) g L⁻¹. (Na = 22.990, Cl = 35.453) 4 MARKS
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7. A chemist needs to deliver 0.0500 mol of H₂SO₄ to a reaction. They have a stock solution of 1.00 mol L⁻¹ sulfuric acid. What volume of stock solution should they measure out? Give your answer in mL. 3 MARKS
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8. A pool technician tests a swimming pool and finds that the chlorine concentration is 3.55 g L⁻¹ of Cl₂ (dissolved chlorine gas). The pool holds 50 000 L of water. Calculate the total mass of Cl₂ in the pool, and the number of Cl₂ molecules present. (Cl = 35.453) 5 MARKS
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9. A student prepares a solution by dissolving 5.85 g of NaCl in 100 mL of water. They claim the concentration is 1.00 mol L⁻¹. A second student says: "You made the solution incorrectly — you should have added water to the NaCl to produce exactly 100 mL of solution, not dissolved it in 100 mL of water." Evaluate both the concentration claim and the procedural claim. (Na = 22.990, Cl = 35.453) 5 MARKS
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10. A scientist needs to prepare 500 mL of a 0.500 mol L⁻¹ solution of potassium permanganate (KMnO₄) as a stock solution. They then plan to transfer two 50.0 mL aliquots — one for a titration experiment and one to compare the colour intensity at different concentrations. (a) Calculate the mass of KMnO₄ needed. (b) Outline a complete procedure to prepare the stock solution. (c) Calculate how many moles of KMnO₄ are in each 50.0 mL aliquot. (K = 39.098, Mn = 54.938, O = 15.999) 7 MARKS
Type your answer below:
Answer in your workbook.
NaOH: V = 250÷1000 = 0.250 L; c = 0.500÷0.250 = 2.00 mol L⁻¹
HCl: V = 100÷1000 = 0.100 L; n = c×V = 2.00×0.100 = 0.200 mol
CaCl₂: V = n÷c = 0.0400÷0.160 = 0.250 L = 250 mL
H₂SO₄: V = 600÷1000 = 0.600 L; c = 1.50÷0.600 = 2.50 mol L⁻¹
KMnO₄: V = 50.0÷1000 = 0.0500 L; n = c×V = 0.0200×0.0500 = 1.00×10⁻³ mol
1. C — Must convert 150 mL → 0.150 L first. c = 0.30 ÷ 0.150 = 2.00 mol L⁻¹.
2. B — V = 0.500 L. c = 0.250 ÷ 0.500 = 0.500 mol L⁻¹.
3. A — V = 0.0250 L. n = 0.100 × 0.0250 = 2.50 × 10⁻³ mol.
4. D — MM(NaOH) = 39.997 g mol⁻¹. n = 4.00 ÷ 39.997 = 0.100 mol. V = 0.200 L. c = 0.100 ÷ 0.200 = 0.500 mol L⁻¹.
5. C — c = 18.0 ÷ 180.16 = 0.100 mol L⁻¹.
Q6 (4 marks):
MM(NaCl) = 22.990 + 35.453 = 58.443 g mol⁻¹ n = m ÷ MM = 9.00 ÷ 58.443 = 0.1540 mol (a) c = n ÷ V = 0.1540 ÷ 1.00 = 0.154 mol L⁻¹ (b) c (g L⁻¹) = 9.00 g ÷ 1.00 L = 9.00 g L⁻¹ (or: 0.154 × 58.443 = 9.00 g L⁻¹)Q7 (3 marks):
V = n ÷ c = 0.0500 ÷ 1.00 = 0.0500 L = 50.0 mLQ8 (5 marks):
Total mass = c × V = 3.55 × 50 000 = 177 500 g = 177.5 kg MM(Cl₂) = 2 × 35.453 = 70.906 g mol⁻¹ n = m ÷ MM = 177 500 ÷ 70.906 = 2503 mol N = n × Nₐ = 2503 × 6.022 × 10²³ = 1.507 × 10²⁷ molecules6. D — MM(NaCl) = 22.990 + 35.453 = 58.443 g mol⁻¹. m for 0.250 mol = 0.250 × 58.443 = 14.61 g ✓. In 500 mL: c = 0.250 ÷ 0.500 = 0.500 mol L⁻¹. The first student's preparation is valid. The second student's suggestion (250 mL) would give c = 1.00 mol L⁻¹, which is also valid but not what was asked. Neither approach is "wrong".
7. A — n = 0.200 × 0.250 = 0.0500 mol. MM(CuSO₄·5H₂O) = 63.546 + 32.06 + 4(15.999) + 5[2(1.008) + 15.999] = 249.68 g mol⁻¹. m = 0.0500 × 249.68 = 12.48 g. Dissolve, transfer to 250 mL volumetric flask, make up to the mark. Option D is wrong because it ignores the water of crystallisation when weighing out.
Q9 (5 marks):
Concentration check: MM(NaCl) = 58.443 g mol⁻¹ [1]. n = 5.85 ÷ 58.443 = 0.1001 mol [1]. If total volume ≈ 100 mL = 0.100 L: c = 0.1001 ÷ 0.100 = 1.00 mol L⁻¹ [1]. The concentration claim is approximately correct — but only if adding the solid to 100 mL of water does not significantly change the total volume.
Procedural evaluation: The second student is correct [1]. Adding water to a fixed volume of NaCl is wrong for accurate concentration preparation. To make exactly 100 mL of 1.00 mol L⁻¹, dissolve the NaCl in a small volume of water (~50 mL), then transfer to a 100 mL volumetric flask and add distilled water exactly to the 100 mL mark [1].
Q10 (7 marks):
(a) MM(KMnO₄) = 39.098 + 54.938 + 4(15.999) = 158.032 g mol⁻¹ [1] n = c × V = 0.500 × 0.500 = 0.250 mol [1] m = n × MM = 0.250 × 158.032 = 39.51 g [1](b) Procedure: Weigh 39.51 g of KMnO₄ on a balance [1]. Dissolve in ~300 mL of distilled water in a beaker with stirring. Once dissolved, quantitatively transfer to a 500 mL volumetric flask. Rinse the beaker 3× with small volumes of distilled water, adding the rinsings to the flask. Add distilled water to the 500 mL mark, stopper and invert several times to mix [1].
(c) n = c × V = 0.500 × 0.0500 = 0.0250 mol per aliquot [1]Each 50.0 mL aliquot contains the same concentration and 0.0250 mol of KMnO₄ [1].
Climb platforms, hit checkpoints, and answer questions on concentration in mol/L, n=cV calculations and solution preparation. Quick recall from lessons 1–6.
Tick when you've finished all activities and checked your answers.