HSC Exam Practice

Sample response. Molar volume (V_m) is the volume occupied by one mole of any gas at a specified temperature and pressure; units are L mol⁻¹. (i) At STP (0 °C, 100 kPa), V_m = 22.71 L mol⁻¹ (NESA standard). (ii) At SATP (25 °C, 100 kPa), V_m = 24.8 L mol⁻¹.

Marking notes. 1 mark for a correct definition of molar volume including the idea of “one mole at specified conditions”; 1 mark for STP temperature AND pressure AND V_m value all correct; 1 mark for SATP temperature AND pressure AND V_m value all correct; 1 mark for correct units (L mol⁻¹) stated or implied.

Sample response. Avogadro’s law states that equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules. Both 1 mol H₂ and 1 mol SF₆ contain exactly 6.022 × 10²³ molecules. Gas volume is almost entirely empty space between particles; the volume is determined by the number of particle–wall collisions per second, not by the mass or size of individual molecules. Because both samples contain the same number of particles at the same temperature and pressure, they exert the same pressure on their container and therefore occupy the same volume, despite SF₆ being approximately 72 times heavier per molecule.

Marking notes. 1 mark for a correct statement of Avogadro’s law; 1 mark for identifying that both samples contain equal numbers of particles (N_A); 1 mark for explaining that gas volume is dominated by empty space and molecular size is negligible, so volume depends on particle count and conditions, not mass.

Sample response. SATP: V_m = 24.8 L mol⁻¹. V = n × V_m = 3.50 × 24.8 = 86.8 L.

Marking notes. 1 mark for correct formula and substitution with V_m = 24.8 L mol⁻¹; 1 mark for correct final answer 86.8 L. Deduct 1 mark if student uses 22.71 (STP) without justification.

Sample response. (a) SATP: n = V ÷ V_m = 4.96 ÷ 24.8 = 0.200 mol. (b) STP: n = 4.96 ÷ 22.71 = 0.218 mol. The answers differ because STP is at a lower temperature (0 °C vs 25 °C). At lower temperature, gas particles move more slowly and the gas occupies a smaller molar volume, so the same volume of gas contains more moles at STP than at SATP.

Marking notes. 1 mark for correct SATP answer (0.200 mol); 1 mark for correct STP answer (0.218 mol); 1 mark for recognising that the same volume contains more moles at the lower temperature; 1 mark for linking this to the smaller molar volume at STP due to lower temperature.

Sample response. STP is 0 °C and 100 kPa (V_m = 22.71 L mol⁻¹); SATP is 25 °C and 100 kPa (V_m = 24.8 L mol⁻¹). The NSW HSC default is 24.8 L mol⁻¹ (SATP) when conditions are unspecified, because SATP approximates typical laboratory conditions in Australia. The old value of 22.4 L mol⁻¹ is no longer used because it applied to the older definition of STP (0 °C, 1 atm = 101.325 kPa); IUPAC updated the standard pressure to 100 kPa in 1982, which gives the slightly higher value of 22.71 L mol⁻¹ at 0 °C.

Marking notes. 1 mark for correctly distinguishing STP and SATP by temperature, pressure, and molar volume; 1 mark for correctly identifying 24.8 L mol⁻¹ (SATP) as the NSW HSC default with a valid reason; 1 mark for explaining why 22.4 L mol⁻¹ is outdated (old pressure standard 1 atm ≠ 100 kPa).

Sample response (a). The temperature is 0 °C and pressure is 100 kPa, which are STP conditions. Therefore the correct molar volume is V_m = 22.71 L mol⁻¹ [1]. n = V ÷ V_m = 3.72 ÷ 22.71 = 0.1638 mol (accept 0.164 mol) [1]. Award 1 mark for correct identification and justification of V_m; 1 mark for correct calculation of n.

Sample response (b). MM = m ÷ n = 7.44 ÷ 0.1638 = 45.4 g mol⁻¹ (accept 44–46 depending on rounding) [1]. From the table, this is closest to CO₂ (44.01 g mol⁻¹) or C₃H₈ (44.10 g mol⁻¹) [1]. The gas is most likely CO₂ given the context of a laboratory chemical reaction producing a gas; propane would typically not be produced as a reaction product in a standard Year 11 chemistry context [1].

Sample response (c). The molar mass is 44.01 g mol⁻¹ (using CO₂). n = 7.44 ÷ 44.01 = 0.1690 mol. At SATP: V = 0.1690 × 24.8 = 4.19 L [1]. The number of moles would not change because the mass of gas is the same; n = m ÷ MM is independent of temperature and pressure. Only the volume changes (it increases at the higher SATP temperature) [1].

Marking notes. (a) 1 mark: correct V_m (22.71) with STP justification; 1 mark: n ≈ 0.164 mol. (b) 1 mark: MM ≈ 44–46 g mol⁻¹; 1 mark: identifies CO₂ or C₃H₈ from table; 1 mark: justification for preferring CO₂. (c) 1 mark: correct volume at SATP (~4.19 L or consistent with student’s n); 1 mark: explains moles are unchanged (mass unchanged; n = m ÷ MM).

Sample response.

Evaluating Student A: MM(N₂) = 2 × 14.007 = 28.014 g mol⁻¹. Mass for 0.500 mol = 0.500 × 28.014 = 14.007 g, not 14.014 g. Student A made an error: they appear to have doubled the atomic mass of a single nitrogen atom (14.007 × 2 for one molecule) but may have used an incorrect mass of 14.014 instead of 14.007 g mol⁻¹ for nitrogen. The correct mass is 14.007 g, not 14.014 g. Even if the calculation were correct, measuring the mass of a gas on a balance is impractical in a laboratory because gas is invisible and extremely light; any air currents or evaporation errors make mass measurements of gases unreliable without specialised apparatus (sealed container + balance).

Evaluating Student B: V = n × V_m = 0.500 × 24.8 = 12.4 L. Student B is correct: collecting 12.4 L of N₂ at SATP gives exactly 0.500 mol. In practice, measuring a gas volume using a calibrated gas syringe or gas collection apparatus is far more practical and accurate than weighing a gas, because gas volumes can be measured directly with standard laboratory equipment such as syringes, eudiometer tubes, or gas collection over water.

Relative merits — mass vs volume: Measuring mass requires a sealed container on an analytical balance, which is difficult for gases. A small leak invalidates the result. By contrast, measuring volume with a gas syringe at known temperature and pressure is straightforward. The molar volume approach (V = n × V_m) converts directly between volume and moles without needing to weigh the gas. However, volume measurement requires knowing the temperature and pressure precisely, and real gases may deviate from ideal behaviour; a temperature or pressure error will lead to an incorrect n calculation. Mass measurement would be more reliable for non-ideal conditions or very precise work.

Conclusion: Student A’s mass value (14.014 g) contains a small numerical error; the correct mass would be 14.007 g. Student B’s proposal is correct. For practical laboratory work, Student B’s volume-based method is preferred for measuring gas quantities.

Marking criteria (10 marks). 1 = MM(N₂) correctly calculated as 28.014 g mol⁻¹. 1 = correct mass for 0.500 mol identified as 14.007 g. 1 = Student A’s error identified (used 14.014 instead of 14.007) with working shown. 1 = Student B’s volume verified: V = 0.500 × 24.8 = 12.4 L with working shown. 1 = Student B explicitly stated as correct. 1 = practical limitation of mass measurement of gas identified (e.g. requires sealed container, impractical on open balance). 1 = advantage of volume measurement stated (e.g. direct measurement with syringe, widely available apparatus). 1 = limitation of volume measurement identified (e.g. requires accurate T and P; ideal gas assumption). 1 = evaluative comparison integrating both methods (not just listing). 1 = precise chemical terminology and clear logical structure throughout (molar mass, SATP, n = V ÷ V_m, ideal gas).