HSC Exam Practice

Sample response. The mole is the SI unit of amount of substance; one mole contains 6.022 × 10²³ elementary entities (Avogadro’s number, N_A). The mole links macroscopic measurements to atomic scale because it defines a fixed number of particles: by measuring the mass of a sample and dividing by the molar mass (g/mol), chemists can calculate the number of moles and then multiply by N_A to find the actual number of atoms, molecules, or ions present. This conversion makes it possible to weigh out chemically equivalent amounts of different substances on a balance.

Marking notes. 1 mark for defining the mole as the SI unit of amount of substance containing 6.022 × 10²³ entities; 1 mark for stating the value of Avogadro’s number (6.022 × 10²³ mol⁻¹); 1 mark for explaining how the mole connects mass (via molar mass) to particle number (via N_A).

Sample response. At STP (Standard Temperature and Pressure as defined by NESA: 0°C and 100 kPa), the molar volume of any ideal gas is 22.71 L/mol. At RTP (Room Temperature and Pressure: 25°C and 100 kPa), the molar volume is 24.8 L/mol. The difference arises because at higher temperature (25°C vs 0°C) the gas molecules move faster and occupy a larger volume per mole at the same pressure.

Marking notes. 1 mark for NESA STP molar volume = 22.71 L/mol with conditions (0°C, 100 kPa); 1 mark for RTP molar volume = 24.8 L/mol with conditions (25°C, 100 kPa); 1 mark for explaining why the values differ (higher temperature at RTP → greater volume). Do not accept 22.4 L/mol as the STP value (that corresponds to the older 0°C, 1 atm standard).

Sample response. The limiting reagent is the reactant that is completely consumed first and determines the maximum amount of product that can form. If the wrong reactant is used for yield calculations, the theoretical yield will be overestimated. To identify the limiting reagent: (1) convert the mass of each reactant to moles; (2) divide each n value by its stoichiometric coefficient from the balanced equation; (3) the reactant with the smaller n ÷ coefficient ratio is the limiting reagent. All product calculations use only the limiting reagent’s moles.

Marking notes. 1 mark for explaining why the limiting reagent matters (determines max product; using wrong reactant overestimates yield); 1 mark for describing the identification method (convert to moles, divide by coefficient); 1 mark for stating the comparison criterion (smaller n ÷ coefficient value = limiting reagent).

Sample response. Percentage purity measures the proportion of a desired substance in an impure sample; it is applied before stoichiometry to find the mass of pure reactant available: m(pure) = m(sample) × (purity ÷ 100). It is less than 100% because real samples contain impurities from their source or prior reactions. Percentage yield measures the proportion of the theoretical maximum product actually obtained; it is calculated after stoichiometry: % yield = (actual yield ÷ theoretical yield) × 100. It is less than 100% because of side reactions, incomplete reactions, or product lost during collection and purification.

Marking notes. 1 mark for correct definition of % purity (proportion of pure substance in sample) and when applied (before stoichiometry); 1 mark for one valid reason % purity < 100% (impurities present from source, manufacture, or storage); 1 mark for correct definition of % yield (actual ÷ theoretical × 100) and when applied (after stoichiometry); 1 mark for one valid reason % yield < 100% (side reactions, loss of product, incomplete reaction, impure reactant).

Sample response. NaOH cannot be used as a primary standard because it is hygroscopic — it absorbs water vapour from the atmosphere — and also reacts with atmospheric CO₂ to form Na₂CO₃. This means the actual mass of pure NaOH in a weighed sample is unknown, so a solution prepared from it does not have a precisely known concentration. One suitable primary standard is anhydrous sodium carbonate (Na₂CO₃), which is stable, non-hygroscopic, and has a large molar mass (105.99 g/mol) that minimises the percentage uncertainty in weighing.

Marking notes. 1 mark for explaining why NaOH fails (hygroscopic / absorbs CO₂); 1 mark for naming a valid primary standard (Na₂CO₃, potassium hydrogen phthalate KHP, anhydrous oxalic acid, or potassium iodate KIO₃); 1 mark for one property making it suitable (stable, non-hygroscopic, high purity, large molar mass, accurately known composition).

Sample response. Step 1 — Balance the equation: common error is changing subscripts inside formulas instead of coefficients. Step 2 — Convert given mass to moles using n = m ÷ MM: common error is using the molar mass of the product instead of the given reactant. Step 3 — Apply the mole ratio from the balanced equation: n(wanted) = n(given) × (coefficient of wanted ÷ coefficient of given): common error is assuming a 1:1 ratio when the coefficients are unequal. Step 4 — Convert moles of product to the required unit (mass, volume, concentration): common error is using the wrong molar mass (using the reactant’s MM instead of the product’s MM, or forgetting to multiply by molar volume for gases).

Marking notes. 1 mark per step correctly described with its most common error identified (4 marks total). Accept variations in wording provided the step and error are clearly matched.

Part (a) — Limiting reagent (3 marks). n(Pb(NO₃)₂) = 0.250 × 0.0750 = 0.01875 mol; ÷ coefficient 1 = 0.01875 [1]. n(KI) = 0.350 × 0.0750 = 0.02625 mol; ÷ coefficient 2 = 0.01313 [1]. KI has the smaller ratio (0.01313 < 0.01875), therefore KI is the limiting reagent; Pb(NO₃)₂ is in excess [1].

Part (b) — Mass of PbI₂ (2 marks). Ratio KI : PbI₂ = 2 : 1; n(PbI₂) = 0.02625 ÷ 2 = 0.01313 mol [1]. MM(PbI₂) = 207.2 + 2(126.90) = 461.0 g/mol; m(PbI₂) = 0.01313 × 461.0 = 6.052 ≈ 6.05 g [1].

Part (c) — Concentration of excess reagent (3 marks). n(Pb(NO₃)₂) consumed = n(KI) × (1 ÷ 2) = 0.02625 × 0.5 = 0.01313 mol [1]. n(Pb(NO₃)₂) remaining = 0.01875 − 0.01313 = 0.005625 mol [1]. V(total) = 75.0 + 75.0 = 150.0 mL = 0.1500 L; c(Pb(NO₃)₂) = 0.005625 ÷ 0.1500 = 0.03750 mol/L [1].

Part (d) — Assumption (1 mark). The assumption is that the volumes of the two solutions are additive (V_total = V₁ + V₂). In reality, mixing solutions of different compositions can cause a slight volume contraction or expansion, so the actual total volume may differ slightly from 150.0 mL, introducing a small error in the calculated concentration [1].

Part (a) — c(Na₂CO₃) (2 marks). n(Na₂CO₃) = 0.530 ÷ 105.99 = 5.001 × 10⁻³ mol [1]. c(Na₂CO₃) = 5.001 × 10⁻³ ÷ 0.2500 = 0.02000 mol/L [1].

Part (b) — Concordant titres and average (2 marks). Rough (19.3 mL) is excluded; T1 (20.85), T2 (20.90), T3 (20.80) are all within 0.10 mL of each other and are concordant [1]. Average = (20.85 + 20.90 + 20.80) ÷ 3 = 20.85 mL. The rough titre is excluded because it was performed quickly to locate the endpoint and the result is unreliable due to the risk of overshooting [1].

Part (c) — c(HCl) (3 marks). n(Na₂CO₃ in aliquot) = 0.02000 × 0.02500 = 5.000 × 10⁻⁴ mol [1]. Ratio Na₂CO₃ : HCl = 1 : 2; n(HCl) = 5.000 × 10⁻⁴ × 2 = 1.000 × 10⁻³ mol [1]. c(HCl) = 1.000 × 10⁻³ ÷ 0.02085 = 0.04796 ≈ 0.0480 mol/L [1].

Sample response. The mole is the central concept of Module 2 because every quantitative calculation — whether involving mass, particle count, gas volume, or solution concentration — must pass through moles as an intermediate step. The four fundamental conversions illustrate this: mass to moles (n = m ÷ MM), particles to moles (n = N ÷ N_A), gas volume to moles (n = V ÷ V_m), and solution to moles (n = cV). None of these quantities can be directly compared or converted into each other without first expressing them in moles. For mass–mass stoichiometry, for example, the four-step method makes this explicit: convert given mass to moles, apply the mole ratio from the balanced equation, and convert product moles to mass. Skipping step 2 (assuming a 1:1 ratio) is the most common error, and it arises precisely from not understanding the mole ratio. For titration back-calculations, the same logic applies: n(standard) = c × V(titre) → mole ratio → n(unknown) → c(unknown) = n ÷ V(flask). The flask volume and titre volume are both used, but at different points — confusing them produces an incorrect concentration because the mole is the bridge between the two. Incorrect application of the mole concept cascades through the calculation: if molar mass is used incorrectly in step 2, the wrong number of moles is carried forward, the ratio step gives an erroneous n(product), and the final mass or concentration is proportionally wrong. Similarly, failing to convert mL to L before using c = n ÷ V introduces a factor-of-1000 error in the calculated concentration. The mole also enables the concept of limiting reagent: only by comparing n ÷ coefficient for each reactant in moles can the analyst determine which substance stops the reaction. Without the mole framework, there is no chemically meaningful way to compare different substances directly. In summary, the claim is justified: the mole is not merely one formula among many in Module 2 — it is the dimensionless currency through which all quantities in the module are converted, compared, and combined.

Marking criteria (8 marks). 1 = explains the four conversions into/out of moles with correct formulae (n = m/MM; n = N/N_A; n = V/V_m; n = cV). 1 = analyses how mass–mass stoichiometry relies on moles as the intermediary, with reference to the mole ratio from the balanced equation. 1 = analyses a second specific calculation type correctly (titration back-calculation, limiting reagent, dilution, gas stoichiometry, or % yield/purity). 1 = discusses the consequences of an error at the mole conversion step (e.g. wrong molar mass, mL instead of L, wrong molar volume) with a specific example of the downstream impact. 1 = explains how the stoichiometric mole ratio enables quantitative prediction of reaction outcomes. 1 = discusses the limiting reagent concept and links it to the mole framework (n ÷ coefficient comparison). 1 = integrates at least two named Module 2 calculation types explicitly. 1 = reaches an explicit, well-reasoned evaluative conclusion that supports or qualifies the claim using chemical evidence.