HSC Exam Practice

Sample response. A mole is the SI unit of amount of substance; it is the amount of substance containing exactly 6.022 × 10²³ elementary entities (atoms, molecules, ions, etc.); the unit is mol. The molar mass is the mass of one mole of a substance, obtained by summing the atomic masses of all atoms in the formula from the periodic table; the unit is g mol⁻¹.

Marking notes. 1 mark for correct definition of mole including the unit mol; 1 mark for stating 6.022 × 10²³ entities per mole (Avogadro’s number); 1 mark for correct definition of molar mass with unit g mol⁻¹. Deduct if units are missing or incorrect.

Sample response. (a) STP: 0 °C and 100 kPa; V_m = 22.71 L mol⁻¹. (b) SATP: 25 °C and 100 kPa; V_m = 24.8 L mol⁻¹. The two values differ because gas volume is temperature-dependent (Charles’s Law): at higher temperature, gas particles have greater average kinetic energy and move further apart, so the same number of moles occupies a larger volume. SATP is 25 °C higher than STP, producing a larger molar volume.

Marking notes. 1 mark per correct V_m value with its full conditions (temperature and pressure); 1 mark for each correct value (2 marks). 1 mark for a correct explanation of why the values differ (temperature/kinetic energy / Charles’s Law). 1 mark for explicitly linking temperature increase to volume increase.

Sample response. An empirical formula gives the simplest whole-number ratio of atoms in a compound; a molecular formula gives the actual number of each type of atom in one molecule. Example where they are identical: water, H₂O — the simplest ratio and the actual molecular formula are both H₂O. Example where they differ: ethene, molecular formula C₂H₄, empirical formula CH₂; glucose, molecular formula C₆H₁₂O₆, empirical formula CH₂O.

Marking notes. 1 mark for correct definition of empirical formula (simplest whole-number ratio); 1 mark for correct definition of molecular formula (actual numbers per molecule); 1 mark for a valid example where EF = MF; 1 mark for a valid example where EF ≠ MF with both formulas stated.

Sample response. Al₂(SO₄)₃ contains: 2 Al, 3 S, and 3 × 4 = 12 O. MM = 2(26.98) + 3(32.06) + 12(16.00) = 53.96 + 96.18 + 192.00 = 342.14 g mol⁻¹.

Marking notes. 1 mark for correctly expanding brackets (3 S and 12 O); 1 mark for correct substitution and arithmetic; 1 mark for the correct final answer with unit. A common error is treating the formula as 2 Al + 1 S + 4 O (342.14 vs 185.96 — wrong answer gets 0 for the final mark but may still receive the brackets mark if working is shown).

Sample response. Rounding intermediate values introduces rounding error at each step; these errors compound through the calculation and can make the final answer significantly wrong even though the method is correct [1 mark for identifying the problem]. Example: n = 15 ÷ 18.015 = 0.8327 mol; if rounded to 0.83, then V = 0.83 × 24.8 = 20.58 L, whereas the correct answer is V = 0.8327 × 24.8 = 20.65 L — a 0.07 L error from one early rounding [1 mark for quantified example]. The student should carry full calculator precision through all intermediate steps and round only the final answer to an appropriate number of significant figures [1 mark for correct practice].

Sample response. There is no direct formula linking gas volume to number of particles, because moles is the central “currency” that connects all four pathways; no pathway skips it [1 mark for justification]. Step 1: use n = V ÷ V_m (L04 gas volume pathway) to convert volume to moles [1 mark]. Step 2: use N = n × N_A (L01 particles pathway) to convert moles to number of particles [1 mark].

Sample response (a) — Trial 1 working. Reaction: 2H₂O → 2H₂ + O₂; so n(O₂) = n(H₂O) ÷ 2. n(H₂O) = m ÷ MM = 0.180 ÷ 18.015 = 9.992 × 10⁻³ mol. n(O₂) = 9.992 × 10⁻³ ÷ 2 = 4.996 × 10⁻³ mol. V_m = V(O₂) ÷ n(O₂) = 0.124 ÷ 4.996 × 10⁻³ = 24.82 L mol⁻¹. Trial 2: n(O₂) = 9.991 × 10⁻³ mol; V_m = 0.248 ÷ 9.991 × 10⁻³ = 24.82 L mol⁻¹. Trial 3: n(O₂) = 1.499 × 10⁻² mol; V_m = 0.371 ÷ 1.499 × 10⁻² = 24.75 L mol⁻¹. Marking: 1 mark per completed row (3 marks for n column; 2 marks for V_m column first two rows; accept ±0.05 rounding). Full working for Trial 1: 2 additional marks (1 formula; 1 correct answer).

Sample response (b). Average V_m = (24.82 + 24.82 + 24.75) ÷ 3 = 24.80 L mol⁻¹. Percentage error = |24.80 − 24.8| ÷ 24.8 × 100 = 0%. Excellent agreement in this case. One source of discrepancy in a real experiment: water vapour pressure in the gas syringe would inflate the measured volume; or temperature slightly above/below 25 °C affects V_m. Marking: 1 mark average; 1 mark percentage error with correct formula; 1 mark for a valid, explained source of discrepancy.

Sample response (c). Smaller volume [1]. At 10 °C, gas molecules have lower average kinetic energy and move less vigorously, so they occupy less volume for the same number of moles (Charles’s Law: V ∝ T at constant pressure). The molar volume at 10 °C < 24.8 L mol⁻¹, so the same number of moles of O₂ produced would occupy a smaller volume than 124 mL [1].

Sample response. Moles are described as the “central currency” because they are the only quantity that connects all other quantities in IQ1. There is no formula that converts, for example, mass directly to a number of particles, or gas volume directly to mass — every pathway must pass through moles as an intermediate step. This is because moles represent an absolute count of entities scaled by Avogadro’s number, and it is this count that relates to mass via the molar mass (a substance-specific constant from the periodic table) or to volume via the molar volume (a condition-specific constant). Skipping the moles step means using the wrong formula for the conversion and produces dimensionally inconsistent answers: multiplying grams by Avogadro’s number, for example, produces units of g entities mol⁻¹, not a pure count of particles — the answer is numerically meaningless. In Australian industrial context: at the Kwinana nickel refinery in Western Australia, engineers must calculate precisely how many kilograms of nickel sulfate (NiSO₄, MM = 154.75 g mol⁻¹) to produce per day from a target output of 5.00 × 10²⁵ Ni atoms. Step 1 (L01): n = N ÷ N_A = 5.00 × 10²⁵ ÷ 6.022 × 10²³ = 83.04 mol. Step 2 (L02): m = n × MM = 83.04 × 154.75 = 12 850 g (12.85 kg). If the moles step were skipped and the engineer multiplied 5.00 × 10²⁵ × 154.75 directly, the result (7.74 × 10²⁷ g) is physically nonsensical and 10²⁴-fold too large. Both pathways (L01 and L02) must contribute; neither alone can complete the conversion. Moles therefore function as the universal, dimensionally correct “exchange rate” between all measured quantities in quantitative chemistry.

Marking criteria (6 marks). 1 = explains why moles are the central currency (no direct formula exists between any two non-moles quantities; pathway must pass through moles). 1 = identifies at least two of the four IQ1 pathways and the quantities they connect. 1 = evaluates the consequence of skipping moles with a specific example (dimensional/unit inconsistency or numerical error). 1 = a named real or Australian industrial context introduced correctly. 1 = at least two pathways applied correctly with full working in the chosen example. 1 = reaches an explicit evaluative conclusion integrating why the moles-as-currency model is chemically necessary and not just a convention.