HSC Exam Practice

Sample response. Percentage purity is the ratio of the mass of the pure desired substance to the total mass of the sample, expressed as a percentage. Formula: % purity = (m_pure ÷ m_sample) × 100.

Marking notes. 1 mark for a correct definition referencing both pure and total mass; 1 mark for the correct formula with symbols or in words.

Sample response. When the total sample mass is used in n = m ÷ MM, the calculation assumes all of the mass is the desired substance. Because the impure sample includes some mass from contaminants, using the total mass artificially inflates the calculated moles of the solute. Since c = n ÷ V, a higher n gives a higher calculated c. The actual moles of the desired substance are less than the total moles calculated; therefore the true concentration is lower than what is reported.

Marking notes. 1 mark for identifying that total mass includes impurity mass, which is not the solute of interest; 1 mark for explaining that this inflates n = m/MM; 1 mark for linking the inflated n to an overestimated c = n/V.

Sample response. Step 1: Apply purity to find m(pure) = m(sample) × (% purity ÷ 100). Step 2: Calculate moles using n = m(pure) ÷ MM (volume must be in litres). Step 3: Calculate concentration using c = n ÷ V.

Marking notes. 1 mark per correctly placed step in the correct order. Accept any equivalent wording. Deduct marks if purity is applied after the n = m/MM step.

Sample response. To convert mg L⁻¹ to mol L⁻¹: (1) Divide the mg L⁻¹ value by 1000 to convert to g L⁻¹. (2) Divide the g L⁻¹ value by the molar mass of the species (in g mol⁻¹) to obtain mol L⁻¹. The information required is the molar mass of the dissolved species.

Marking notes. 1 mark for the ÷ 1000 (mg to g) step; 1 mark for the ÷ MM step; 1 mark for stating the molar mass is the required information.

Sample response. In a forward calculation, you start with m(sample) and multiply by (% purity ÷ 100) to obtain m(pure), then proceed m(pure) → n → c. Purity reduces the mass used in subsequent steps. In a back-calculation, you start with the target concentration, calculate n = c × V, then m(pure) = n × MM. Purity is applied last, dividing m(pure) by (% purity ÷ 100) to obtain the larger m(sample) that must be weighed out.

Marking notes. 1 mark for correctly describing the forward direction (m(sample) → m(pure) via purity, then to n and c); 1 mark for correctly describing the back-calculation direction (target c → n → m(pure) → m(sample) by dividing by purity); 1 mark for explicitly stating purity is a multiplier in forward calculations; 1 mark for explicitly stating purity is a divisor in back-calculations (producing a larger result).

Sample response. 0.9% w/v = 0.9 g per 100 mL = 9.0 g per 1000 mL = 9.0 g L⁻¹. [1] Converting to mol L⁻¹: c = 9.0 ÷ 58.44 = 0.1540 mol L⁻¹ ≈ 0.154 mol L⁻¹. [1] This confirms the equivalence; the saline is isotonic with blood plasma because it approximates the physiological NaCl concentration. [1]

Marking notes. 1 mark for converting 0.9% w/v to 9.0 g L⁻¹; 1 mark for correctly dividing by MM to get 0.154 mol/L; 1 mark for a valid confirmatory statement or context (e.g. isotonic, physiological).

Part (a). Concentration per 5.0 mL dose: 250 mg in 5.0 mL = 50 000 mg per 1000 mL = 50 g L⁻¹. [1] c = 50.0 g L⁻¹ ÷ 151.16 g mol⁻¹ = 0.3309 mol L⁻¹. [1]

Part (b). n = c × V = 0.3309 × 200 = 66.18 mol. [1] m(pure) = n × MM = 66.18 × 151.16 = 10 004 g ≈ 10.0 kg. [1] (Accept also: 250 mg/5.0 mL = 50 g/L; 50 g/L × 200 L = 10 000 g = 10.0 kg directly.)

Part (c). m(sample) = m(pure) ÷ (purity ÷ 100) = 10 004 ÷ 0.992 = 10 084 g ≈ 10.08 kg. [1] (Only ~84 g extra is needed due to the very high 99.2% purity.) [1 for correct method applied]

Part (d). Target concentration = 0.3309 mol/L; measured = 0.328 mol/L. % difference = |(0.3309 − 0.328)| ÷ 0.3309 × 100 = 0.0029 ÷ 0.3309 × 100 = 0.88%. [1] This is within an acceptable analytical tolerance (typically <1%). [1] One non-weighing reason: incomplete dissolution of paracetamol powder due to inadequate mixing, leaving some undissolved paracetamol that was not measured in the sample tested. Alternatively: the volume of suspension prepared was slightly greater than 200 L (dilution error), reducing the concentration below target. [1 for any valid, specific, non-weighing reason]

Marking notes. (a) 1 = correct conversion of 250 mg in 5 mL to g/L; 1 = correct c in mol/L. (b) 1 = correct n or equivalent direct method; 1 = correct m(pure) ~10.0 kg. (c) 1 = correct method (divide by purity); 1 = correct m(sample) ~10.08 kg. (d) 1 = correct % difference (allow ±0.2%); 1 = valid evaluation of whether this is acceptable; 1 = one specific non-weighing cause of discrepancy.

Sample response. Percentage purity is a critical parameter whenever a chemical reagent is obtained from a commercial supplier, as virtually all industrial-grade and even many reagent-grade chemicals contain small amounts of contaminants. Failing to account for purity introduces a systematic error that affects every calculation in a chain, with real consequences that scale with the context. In a forward calculation (finding c from a weighed impure sample), ignoring purity means the calculated moles n = m(sample) ÷ MM exceeds the true moles of solute, giving an overestimated concentration. For example, if a 96% pure NaOH sample is assumed to be pure, the concentration would be overestimated by ~4%. In a back-calculation (finding how much impure sample to weigh out), ignoring purity and using only m(pure) — instead of the larger m(sample) = m(pure) ÷ purity — means too little material is weighed, producing a solution below the target concentration. In analytical chemistry (e.g. preparing a standard solution at CSIRO or a university laboratory), an overestimated concentration would cause systematic errors in every titration performed using that standard: the endpoint would be reached with less titrant than expected, and all analyte concentrations derived from the titration would be wrong in the same direction. Over many experiments, this propagates as a systematic bias. In industrial manufacturing (e.g. a pharmaceutical company in Sydney preparing paracetamol suspension for children), an underdose due to ignoring purity means patients receive less active ingredient per dose than prescribed. For a paediatric medication, this could lead to ineffective pain relief; for a medication with a narrow therapeutic window (e.g. digoxin), it could be life-threatening. In environmental monitoring (e.g. a water authority preparing calibration standards for fluoride analysis under the ADWG), a 10% error in purity-adjusted concentration means all fluoride readings from the instrument will be 10% off, potentially leading to miscommunication about whether water meets the 1.50 mg/L WHO guideline. In summary, percentage purity is not a minor correction: it directly determines the accuracy of any mole-based calculation, and the magnitude of the error equals the deviation of purity from 100%. In high-stakes contexts — medicine, food safety, environmental compliance — such errors can have serious public health consequences. Purity must always be applied before n = m/MM, regardless of which direction the calculation runs.

Marking criteria (7 marks). 1 = explains mathematical consequence of ignoring purity in a forward calculation (overestimated n and c). 1 = explains mathematical consequence in a back-calculation (too little sample weighed, under-concentration). 1 = first named Australian context with specific consequence (CSIRO/university analytical lab: titration systematic error; pharmaceutical factory: underdose). 1 = second named Australian context with specific consequence (water authority: fluoride monitoring error; industrial reagent preparation). 1 = links the magnitude of error to the deviation of purity from 100% (i.e. 8% impurity → 8% error in concentration). 1 = distinguishes between forward and back-calculation consequences clearly with correct directional reasoning (over vs under). 1 = reaches a synthesising evaluative judgement on why purity is important, integrating multiple contexts into an overall assessment.