You can do every concentration calculation perfectly on paper and still fail an exam question — because the numbers are wrapped in a real-world scenario you don't recognise. This lesson teaches you to strip away the context, find the numbers, and apply the formulas you already know. That's the skill that separates 70% students from 90% students.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A water quality report states that a river contains lead (Pb) at 0.025 mg L⁻¹ and the safe limit is 0.010 mg L⁻¹. A scientist wants to compare this to a published molar concentration limit. What do you think they would need to do to convert between mg L⁻¹ and mol L⁻¹? What information would they need?
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📚 Core Content
Wrong: Gases have no mass because they float.
Right: Gases have mass; their density is just much lower than solids and liquids.
Real-world chemicals are rarely 100% pure. A bag of "sodium hydroxide" from a supply company might be 97% NaOH with 3% Na₂CO₃ (absorbed from air) and trace water. If you calculate moles assuming 100% purity, your answer will be wrong — and in an industrial or medical context, that error has consequences.
Percentage purity tells you what fraction of a sample is actually the substance you want:
When purity is involved, you must apply it before converting to moles. The workflow is:
HSC questions often use realistic scenarios to wrap the same calculation in unfamiliar language. These examples show you the range of contexts you might see:
Normal saline for intravenous infusion is 0.9% NaCl (by mass/volume — 9 g per 1000 mL). This is isotonic with blood plasma.
≈ 0.154 mol L⁻¹ NaClSafe pool chlorine levels are 1–3 ppm (mg/L) of free Cl₂. Above 5 ppm is irritating; below 0.5 ppm allows bacteria to grow.
≈ 1.4–4.2 × 10⁻⁵ mol L⁻¹Average ocean salinity is 35 g per kilogram of seawater (3.5% by mass). Composed mainly of NaCl with smaller amounts of MgCl₂, MgSO₄.
≈ 0.60 mol L⁻¹ NaClChildren's paracetamol suspension contains 250 mg of paracetamol per 5 mL. Dosing is weight-based to avoid overdose.
≈ 0.332 mol L⁻¹ (MM = 151.16 g mol⁻¹)🧮 Worked Examples
🧪 Activities
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| Contaminant | Formula | MM (g mol⁻¹) | Limit (mg L⁻¹) | Limit (mol L⁻¹) — calculate |
|---|---|---|---|---|
| Nitrate | NO₃⁻ | 62.00 | 50.0 | ? |
| Fluoride | F⁻ | 19.00 | 1.50 | ? |
| Lead | Pb²⁺ | 207.20 | 0.010 | ? |
| Arsenic | As | 74.92 | 0.010 | ? |
Complete the table calculations and answer the questions below:
Complete in your workbook.
At the start of this lesson, you thought about converting lead concentration from mg L⁻¹ to mol L⁻¹ and what information you would need.
The answer: convert mg to g (÷ 1000), then divide by the molar mass (MM) of lead. For Pb: 0.025 mg/L = 2.5 × 10⁻⁵ g/L; c = 2.5 × 10⁻⁵ ÷ 207.2 = 1.2 × 10⁻⁷ mol L⁻¹. The safe limit (0.010 mg/L) = 4.8 × 10⁻⁸ mol L⁻¹. The river exceeds the limit in both units. You needed: the molar mass of Pb, the unit conversion mg→g, and the formula c = m/V ÷ MM.
Reflect: how complete was your initial approach?
Write a reflection in your workbook.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. A laboratory technician has a supply of sulfuric acid (H₂SO₄) that is labelled 98.0% pure (by mass). The solution has a density of 1.84 g mL⁻¹. (a) Calculate the mass of H₂SO₄ in 10.0 mL of this concentrated acid. (b) Calculate the number of moles of H₂SO₄ in 10.0 mL. (H = 1.008, S = 32.06, O = 15.999) 4 MARKS
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7. A river water sample is tested and found to contain chloride ions (Cl⁻) at 250 mg L⁻¹. The river feeds into a reservoir with a capacity of 5.00 × 10⁸ L. (a) Express the chloride concentration in mol L⁻¹. (b) Calculate the total mass of Cl⁻ in the reservoir in kilograms. (Cl = 35.453) 4 MARKS
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8. A student analyses a sample of impure magnesium oxide (MgO). They dissolve 2.50 g of the sample in excess acid and determine that the solution contains 0.0580 mol of Mg²⁺ ions. (a) Calculate the mass of pure MgO in the sample. (b) Calculate the percentage purity of the sample. (Mg = 24.305, O = 15.999) 4 MARKS
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9. A student analyses a sample of impure sodium carbonate (Na₂CO₃) and reports: "I dissolved 5.30 g of the sample in water. I then determined that the solution contains 0.0480 mol of Na⁺ ions. So the purity of Na₂CO₃ is 96.2%." Evaluate the student's working — is their purity calculation correct? (Na = 22.990, C = 12.011, O = 15.999) 5 MARKS
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10. A water treatment plant must reduce the fluoride (F⁻) concentration in drinking water from 3.20 mg L⁻¹ to the WHO guideline of 1.50 mg L⁻¹ by adding calcium hydroxide (Ca(OH)₂) to precipitate CaF₂. The plant processes 2.00 × 10⁶ L of water per day. Design a procedure including: (a) the mass of F⁻ to be removed per day, (b) the moles of F⁻ to be removed, (c) the moles of Ca(OH)₂ required using the reaction Ca(OH)₂ + 2F⁻ → CaF₂ + 2OH⁻, and (d) the mass of Ca(OH)₂ needed. (F = 18.998, Ca = 40.078, O = 15.999, H = 1.008) 7 MARKS
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Scenario A:
n(NaCl) = 3.51 ÷ 58.443 = 0.06006 mol V = 0.500 L; c = 0.06006 ÷ 0.500 = 0.120 mol L⁻¹0.120 mol L⁻¹ is less concentrated than physiological saline (0.154 mol L⁻¹) — this is a hypotonic rehydration solution.
Scenario B:
n = c × V = 0.500 × 2.00 = 1.00 mol m(pure) = 1.00 × 100.09 = 100.1 g m(sample) = 100.1 ÷ 0.880 = 113.7 g ≈ 114 g(i) Highest molar limit: NO₃⁻. This doesn't mean it's least dangerous — molar concentration depends on MM. Nitrate has a higher limit because it is less acutely toxic than lead or arsenic at equivalent mass concentrations.
(ii) Pb²⁺: 0.030 mg/L = 3.0 × 10⁻⁵ g/L; c = 3.0 × 10⁻⁵ ÷ 207.20 = 1.45 × 10⁻⁷ mol L⁻¹This exceeds the ADWG limit of 4.83 × 10⁻⁸ mol L⁻¹ — the water is unsafe.
1. B — m(pure) = 8.00 × 0.940 = 7.52 g.
2. C — m(pure) = 10.0 × 0.900 = 9.00 g. n = 9.00 ÷ 39.997 = 0.225 mol. c = 0.225 ÷ 0.250 = 0.900 mol L⁻¹. A (0.450) is wrong — uses 500 mL instead of 250 mL. B (1.00) is wrong — ignores the 90% purity step. D (0.800) is wrong — uses 80% purity by error.
3. D — 2.00 mg/L = 2.00 × 10⁻³ g/L. c = 2.00 × 10⁻³ ÷ 50.0 = 4.00 × 10⁻⁵ mol L⁻¹.
4. A — m(pure) = 0.300 × 100.09 = 30.03 g. m(sample) = 30.03 ÷ 0.750 = 40.0 g.
5. B — 0.9 g per 100 mL = 9.0 g L⁻¹. c = 9.0 ÷ 58.44 = 0.154 mol L⁻¹.
Q6 (4 marks):
(a) mass of 10.0 mL = density × V = 1.84 × 10.0 = 18.4 g m(pure H₂SO₄) = 18.4 × 0.980 = 18.03 g (b) MM(H₂SO₄) = 2(1.008) + 32.06 + 4(15.999) = 98.072 g mol⁻¹ n = 18.03 ÷ 98.072 = 0.184 molQ7 (4 marks):
(a) 250 mg/L = 0.250 g/L; c = 0.250 ÷ 35.453 = 7.05 × 10⁻³ mol L⁻¹ (b) Total mass = 0.250 g/L × 5.00 × 10⁸ L = 1.25 × 10⁸ g = 1.25 × 10⁵ kgQ8 (4 marks):
(a) MM(MgO) = 24.305 + 15.999 = 40.304 g mol⁻¹ m(MgO) = n × MM = 0.0580 × 40.304 = 2.338 g (b) % purity = (2.338 ÷ 2.50) × 100 = 93.5%6. D — The molar calculation: 0.015 mg/L = 1.5 × 10⁻⁵ g/L; c = 1.5 × 10⁻⁵ ÷ 74.922 = 2.0 × 10⁻⁷ mol L⁻¹ is correct. The conclusion that 0.015 > 0.010 mg/L means the water exceeds the WHO guideline is also correct. The first student is correct in both claims. Option D correctly identifies this — option C incorrectly states the student made an error elsewhere.
7. A — Target NO₃⁻ = 50.0 mg/L in 500 mL = 25.0 mg total. n(NO₃⁻) = 0.0250 ÷ 62.004 = 4.03 × 10⁻⁴ mol. n(KNO₃) = n(NO₃⁻) = 4.03 × 10⁻⁴ mol (1:1). m(KNO₃) = 4.03 × 10⁻⁴ × 101.10 = 0.0408 g = 40.8 mg ✓.
Q9 (5 marks):
MM(Na₂CO₃) = 2(22.990) + 12.011 + 3(15.999) = 105.988 g mol⁻¹ [1]Each Na₂CO₃ produces 2 Na⁺ → n(Na₂CO₃) = n(Na⁺) ÷ 2 = 0.0480 ÷ 2 = 0.0240 mol [1]
m(Na₂CO₃) = 0.0240 × 105.988 = 2.544 g [1] % purity = (2.544 ÷ 5.30) × 100 = 48.0% [1]The student's answer of 96.2% is wrong [1]. They likely forgot that each Na₂CO₃ yields 2 Na⁺ ions and used n(Na₂CO₃) = n(Na⁺) = 0.0480 mol instead of 0.0240 mol. This error doubled their calculated n(Na₂CO₃) and doubled their purity.
Q10 (7 marks):
(a) Concentration to remove = 3.20 − 1.50 = 1.70 mg/L [1] Total mass of F⁻ = 1.70 × 10⁻³ g/L × 2.00 × 10⁶ L = 3400 g = 3.40 kg [1] (b) n(F⁻) = 3400 ÷ 18.998 = 178.97 mol ≈ 179 mol [1](c) Reaction: 1 mol Ca(OH)₂ reacts with 2 mol F⁻ [1]
n(Ca(OH)₂) = 179 ÷ 2 = 89.5 mol [1] (d) MM(Ca(OH)₂) = 40.078 + 2(15.999) + 2(1.008) = 74.092 g mol⁻¹ [1] m(Ca(OH)₂) = 89.5 × 74.092 = 6631 g ≈ 6.63 kg per day [1]Defend your ship by blasting the correct answers for Concentration in Context. Scores count toward the Asteroid Blaster leaderboard.
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