Chemistry • Year 11 • Module 2 • Lesson 8
Concentration in Context
Lock in the key vocabulary, the purity formulas, and the correct calculation sequence before tackling harder contextual problems.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: percentage purity, pure substance, impure sample, m(pure), m(sample), ppm, ppb, mg L−1, back-calculation, % w/v. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The ratio of the mass of the desired substance to the total mass of the sample, expressed as a percentage. | |
| 1.2 | A substance that contains only one chemical species with no contaminants. | |
| 1.3 | A sample that contains the desired substance together with one or more contaminants. | |
| 1.4 | The mass of only the desired chemical species within a sample, after accounting for impurities. | |
| 1.5 | The total mass of a weighed sample, including impurities. | |
| 1.6 | Parts per million; for dilute aqueous solutions, numerically equal to 1 milligram per litre. | |
| 1.7 | Parts per billion; a unit of concentration equal to 1 microgram per litre (µg L−1) in dilute aqueous solutions. | |
| 1.8 | A concentration unit expressing milligrams of solute dissolved per litre of solution. | |
| 1.9 | The process of working from a target concentration backwards to find the mass of sample that must be weighed out. | |
| 1.10 | A concentration expression where the number of grams of solute per 100 mL of solution is stated; 1% w/v = 10 g L−1. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 When a sample is labelled “95% pure”, the mass of the pure substance equals the total sample mass. T / F
2.2 Purity must be applied before converting mass to moles in any calculation involving an impure sample. T / F
2.3 To convert a concentration from mg L−1 to mol L−1, you multiply by the molar mass. T / F
2.4 When back-calculating the mass of an impure sample needed, the mass of impure sample is always greater than the mass of pure substance required. T / F
2.5 A concentration of 0.9% NaCl (w/v) means there are 9 g of NaCl per 1000 mL of solution, which equals 9 g L−1. T / F
2.6 The formula for percentage purity is: % purity = (msample ÷ mpure) × 100. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
purity · molar mass · back-calculate · impurities · larger · moles · litres · divide
Real chemicals from suppliers always contain small amounts of ___________, so you must account for ___________ before calculating moles. The correct order is: first find m(pure) by multiplying the sample mass by the purity fraction, then convert m(pure) to ___________ by dividing by the ___________, then finally divide by the volume in ___________ to get the concentration. To convert a concentration in g L−1 to mol L−1, you ___________ by the molar mass. When a chemist needs to prepare a solution of a specific concentration from an impure reagent, they must ___________ from the target concentration to find the required sample mass. The mass of impure sample required is always ___________ than the mass of pure substance needed.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 What is the purpose of applying the purity fraction to a sample mass, and why must this happen before you use n = m ÷ MM?
4.2 Explain what “ppm” means in a water-quality report and how you convert a reading in mg L−1 to mol L−1.
4.3 A bottle of NaOH is labelled 97% pure. What does this mean for the actual moles of NaOH in a 10.0 g sample weighed on a balance?
4.4 Why is the impure sample mass always greater than the mass of pure substance needed in a back-calculation?
5. Build a concept map
Draw labelled arrows between the seven terms below to show how they connect in a purity calculation. Each arrow must carry a linking phrase (e.g. “is multiplied by”, “divided by”, “is converted to”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: m(sample) · purity fraction · m(pure) · molar mass · n (moles) · volume (L) · concentration (mol L−1).
6. Formula fill-in
Complete each formula below by filling in the missing symbol or expression. 6 marks (1 each)
| # | Formula (fill the blank) | What it calculates |
|---|---|---|
| 6.1 | m(pure) = m(sample) × ( _______ ÷ 100 ) | Mass of pure substance in the sample |
| 6.2 | m(sample) = m(pure) ÷ ( _______ ÷ 100 ) | Sample mass needed for a given pure mass |
| 6.3 | n = _______ ÷ MM | Moles from mass of pure substance |
| 6.4 | c = n ÷ _______ | Molar concentration (V must be in litres) |
| 6.5 | c (mol L−1) = c (g L−1) ÷ _______ | Convert g L−1 to mol L−1 |
| 6.6 | c (g L−1) = c (mg L−1) ÷ _______ | Convert mg L−1 to g L−1 |
Q1 — Term–definition match
1.1 percentage purity • 1.2 pure substance • 1.3 impure sample • 1.4 m(pure) • 1.5 m(sample) • 1.6 ppm • 1.7 ppb • 1.8 mg L−1 • 1.9 back-calculation • 1.10 % w/v.
Q2 — True / false with correction
2.1 False. If the sample is 95% pure, m(pure) = m(sample) × 0.95, which is less than the total sample mass — 5% is impurities.
2.2 True.
2.3 False. To convert mg L−1 to mol L−1, you first convert mg to g (divide by 1000), then divide by the molar mass. Multiplying gives the wrong answer.
2.4 True. Since only a fraction of the sample is the desired pure substance, you always need to weigh out more than the pure mass required. The impure sample is divided by the purity fraction, making it larger.
2.5 True. 0.9 g per 100 mL = 9 g per 1000 mL = 9 g L−1. Correct.
2.6 False. The correct formula is % purity = (mpure ÷ msample) × 100 — the pure mass is on top, the total sample mass is on the bottom.
Q3 — Cloze paragraph
In order: impurities / purity / moles / molar mass / litres / divide / back-calculate / larger.
Q4.1 — Purpose of applying purity first
Applying the purity fraction finds the mass of the actual chemical species (mpure) in the sample. The formula n = m ÷ MM assumes all the mass is the pure substance; using m(sample) directly would overestimate the moles because it includes the mass of the impurities, which do not contribute to the reaction.
Q4.2 — ppm and the mg L−1 to mol L−1 conversion
In a water-quality report, “ppm” means parts per million, numerically equivalent to mg L−1 for dilute aqueous solutions. To convert: first divide mg L−1 by 1000 to get g L−1, then divide by the molar mass (in g mol−1) to get mol L−1.
Q4.3 — 97% pure NaOH
The label means only 97% of the 10.0 g sample is actually NaOH; the remaining 3% is impurities (e.g. Na2CO3, water). So m(NaOH) = 10.0 × 0.97 = 9.70 g, giving n = 9.70 ÷ 39.997 = 0.243 mol — fewer moles than if the sample were 100% pure (0.250 mol).
Q4.4 — Why impure sample mass > pure mass needed
Only a fraction of the impure sample is the desired substance. To get the required amount of pure substance, you must weigh out enough of the mixture such that the fraction (purity ÷ 100) of it equals the pure mass needed. Dividing by a number less than 1 always gives a larger result: m(sample) = m(pure) ÷ (purity ÷ 100) > m(pure).
Q5 — Sample concept map
Correct maps should include arrows such as:
- m(sample) — multiplied by purity fraction gives → m(pure)
- m(pure) — divided by → molar mass — gives → n (moles)
- n (moles) — divided by → volume (L) — gives → concentration
- concentration — multiplied by volume gives → n (moles) (reverse)
- n (moles) — multiplied by molar mass gives → m(pure) (reverse)
- m(pure) — divided by purity fraction gives → m(sample) (reverse)
Award 1 mark per valid labelled arrow (minimum 6).
Q6 — Formula fill-in
6.1 % purity • 6.2 % purity • 6.3 m(pure) • 6.4 V • 6.5 MM (molar mass) • 6.6 1000.