Concentration in Context
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.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
L02 — Molar Mass
$$n = m \div MM$$
Mass (g) → moles
L06 — Concentration
$$c = n \div V$$
V must be in litres
New — Purity
% purity = (mpure ÷ mtotal) × 100
Apply BEFORE calculating moles
Key facts
- % purity formula and what it means
- Realistic concentration ranges for common substances
- How purity adjusts the mass you need to weigh
Concepts
- Why impurities reduce the actual moles present
- How to extract numbers from a contextual problem
- Why you always apply purity before molar mass
Skills
- Calculate mass of pure substance from impure sample
- Solve multi-step problems involving purity + concentration
- Work backwards from concentration to find sample mass
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:
m(pure) = m(sample) × (% purity ÷ 100)
The Calculation Order — Always This Sequence
When purity is involved, you must apply it before converting to moles. The workflow is:
m(pure) = m(sample) × (% purity ÷ 100)
n = m(pure) ÷ MM
c = n ÷ V (V in litres)
Back-calculation (reverse direction): start with target c → multiply by V to get n → multiply by MM to get m(pure) → divide by (purity ÷ 100) to get m(sample) needed.
Concentration in Real Contexts
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). Isotonic with blood plasma. ≈ 0.154 mol L⁻¹ NaCl.
Safe pool chlorine 1–3 ppm (mg/L). Above 5 ppm irritating; below 0.5 ppm bacteria grow. ≈ 1.4–4.2 × 10⁻⁵ mol L⁻¹.
Average ocean salinity 35 g per kilogram (3.5% by mass), mainly NaCl plus MgCl₂, MgSO₄. ≈ 0.60 mol L⁻¹ NaCl.
Children's paracetamol suspension: 250 mg per 5 mL. Dosing is weight-based. ≈ 0.332 mol L⁻¹ (MM = 151.16 g mol⁻¹).
% purity = (mpure ÷ msample) × 100; m(pure) = m(sample) × (% purity ÷ 100). Always apply purity before calculating moles: purity → m(pure) → n = m(pure) ÷ MM → c = n ÷ V. The sample mass is always greater than the pure mass.
Pause — copy the highlighted rule and formula into your book before moving on.
Quick check: A 10.0 g sample of NaOH is 90% pure. Which calculation is correct for moles of NaOH?
Match each concentration term to its definition.
- ppm
- % w/v
- % purity
- ppb
- Mass of pure substance ÷ mass of sample × 100.
- Concentration unit equal to about 1 mg L⁻¹ in dilute aqueous solutions.
- Concentration unit equal to about 1 μg L⁻¹ — used for very low pollutant levels.
- Grams of solute per 100 mL of solution (equivalent to 10 g L⁻¹ for each 1%).
We just saw how percentage purity affects the mass of substance available and must be applied before moles are calculated. That raises a question: how do you handle the variety of concentration units used in real-world contexts (ppm, g L⁻¹, % w/v)? This card answers it → with the conversion rules you need.
Context questions often give concentrations as "mg/L", "g/L", "ppm" or "% w/v". You must recognise these as concentrations and convert.
c (mol L⁻¹) = c (g L⁻¹) ÷ MM. Just divide by the molar mass.
First convert mg → g by ÷ 1000, then ÷ MM. For dilute aqueous solutions, ppm ≈ mg L⁻¹.
% w/v = g per 100 mL = 10 g L⁻¹. So 0.9% w/v NaCl = 9 g L⁻¹.
Concentration unit conversions: g L⁻¹ → mol L⁻¹ by dividing by MM; mg L⁻¹ → g L⁻¹ by ÷ 1000, then ÷ MM; ppm ≈ mg L⁻¹ for dilute aqueous solutions; % w/v = g per 100 mL = 10 g L⁻¹ per 1%.
Add the highlighted conversions to your notes before the check below.
Fill the blanks: drag each token into the correct blank.
To convert mg L⁻¹ to mol L⁻¹: divide by ___ then by ___. 1% w/v equals ___ g L⁻¹. Apply ___ before any moles calc.
Did you get this? True or false: a 0.9% w/v NaCl solution is the same as 9 g L⁻¹.
Worked examples · reveal as you go
A sample of NaOH has a purity of 96.0%. A student dissolves 5.00 g of this impure NaOH in water and makes up to 500 mL of solution. Calculate the actual concentration of NaOH in the solution. (Na = 22.990, O = 15.999, H = 1.008)
A chemist needs to prepare 250 mL of 0.200 mol L⁻¹ NaOH solution. The available NaOH is 95.0% pure. What mass of the impure NaOH must be weighed out? (MM of NaOH = 39.997 g mol⁻¹)
A water sample is found to contain 0.0500 g L⁻¹ of nitrate ions (NO₃⁻). Express this concentration in mol L⁻¹ and determine whether it exceeds the WHO safe drinking limit of 8.06 × 10⁻³ mol L⁻¹. (N = 14.007, O = 15.999)
Common errors · the 3 traps that cost marks
Forgetting to apply purity before calculating moles
The most frequent error. A student reads "5.00 g of NaOH (96% pure)" and immediately writes n = 5.00 ÷ 39.997 — treating the impure sample as if it were all NaOH. The actual moles of NaOH are less, giving an overestimated concentration.
Fix: Every time you see "% pure" or "% purity", underline it and write m(pure) = m(sample) × purity as your first line before anything else.
Dividing by purity instead of multiplying (or vice versa)
When finding how much impure sample to weigh out, students sometimes multiply by purity (giving a smaller number — wrong: you need more impure material than pure). When finding pure mass from sample, they divide (wrong again).
Fix: Think logically. Pure mass < sample mass. Sample → pure: multiply by purity fraction. Pure → sample: divide by purity fraction. The impure sample is always the bigger mass.
Not recognising g L⁻¹ as a concentration unit
Context questions often give concentrations as "mg/L", "g/L", "ppm" or "% w/v". Students sometimes don't recognise these as concentrations and don't know how to convert.
Fix: Any "amount per volume" expression is a concentration. g L⁻¹: c (mol L⁻¹) = c (g L⁻¹) ÷ MM. mg L⁻¹: ÷ 1000 first, then ÷ MM.
A student calculates the concentration of NaOH in a solution made by dissolving 8.00 g of NaOH (95.0% pure) in 500 mL of water. Click the line with the error.
- MM(NaOH) = 22.990 + 15.999 + 1.008 = 39.997 g mol⁻¹
- n = m(sample) ÷ MM = 8.00 ÷ 39.997 = 0.2000 mol
- V = 500 ÷ 1000 = 0.500 L
- c = n ÷ V → final answer reported in mol L⁻¹
Quick-fire practice · 5 reps +2 XP per reveal
Scenario A — Hospital Pharmacy: 3.51 g of NaCl (MM = 58.443 g mol⁻¹) is dissolved in 500 mL of water. Calculate the molar concentration. Compare to physiological saline (0.154 mol L⁻¹).
Scenario B — Industrial CaCO₃ batch is 88.0% pure. Need 2.00 L of 0.500 mol L⁻¹ CaCO₃. (MM = 100.09 g mol⁻¹) Mass of impure CaCO₃ required?
A water sample contains nitrate (NO₃⁻) at 50.0 mg L⁻¹. Convert to mol L⁻¹. (MM(NO₃⁻) = 62.00 g mol⁻¹)
10.0 g of NaOH that is 90.0% pure is dissolved in 250 mL of water. Calculate the molar concentration. (MM = 39.997 g mol⁻¹)
A 0.9% w/v NaCl saline solution. Express in g L⁻¹ and in mol L⁻¹. (MM = 58.44 g mol⁻¹)
Earlier you were asked: How could a scientist convert a lead concentration from mg L⁻¹ to mol L⁻¹?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q1. 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)
Q2. 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)
Q3. 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)
Q4. 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)
Q5. 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)
📖 Comprehensive answers (click to reveal)
Multiple choice — drill bank
MC feedback shown inline as you complete each question. Common items: m(pure) = m(sample) × purity; ppm ≈ mg/L for dilute aqueous solutions; 0.9% w/v = 9 g/L = 0.154 mol/L NaCl; always apply purity BEFORE moles calc.
Short answer model answers
Q1 (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 mol
Q2 (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 kg
Q3 (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%
Q4 (5 marks):
MM(Na₂CO₃) = 2(22.990) + 12.011 + 3(15.999) = 105.988 g mol⁻¹ [1]
Each Na₂CO₃ → 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 purity.
Q5 (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 1:2 (Ca(OH)₂: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]
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