HSC Exam Practice

Sample response. Concentration is the amount of solute (in moles) dissolved per unit volume of solution. Its standard unit in chemistry is mol L⁻¹ (molar; also written M or mol/L).

Marking notes. 1 mark for a correct definition referencing moles per unit volume; 1 mark for stating mol L⁻¹ (accept M or mol/L).

Sample response. The unit mol L⁻¹ requires volume in litres. Substituting 250 mL instead of 0.250 L makes the denominator 1000 times larger, so the calculated concentration will be 1000 times too small (e.g. 0.002 mol L⁻¹ instead of 2.00 mol L⁻¹). The required correction step is to divide the volume in mL by 1000 before substituting: 250 mL ÷ 1000 = 0.250 L.

Marking notes. 1 mark for explaining the unit mismatch (mol L⁻¹ demands litres); 1 mark for explaining the numerical consequence (result is 1000 times too small); 1 mark for stating the correct conversion step (divide mL by 1000).

Sample response. The rearrangement n = c × V calculates the moles of solute in the solution. When c is in mol L⁻¹ and V is in litres, the unit of the answer is mol (moles).

Marking notes. 1 mark for “moles of solute” (accept “amount of solute”); 1 mark for the unit mol.

Sample response. A concentration in mol L⁻¹ expresses the number of moles of solute per litre of solution; it counts particles and is used in all stoichiometric calculations. A concentration in g L⁻¹ expresses the mass of solute in grams per litre of solution; it is convenient when masses are measured directly without needing molar mass. The conversion formula is: c (mol L⁻¹) = c (g L⁻¹) ÷ MM, where MM is the molar mass of the solute in g mol⁻¹.

Marking notes. 1 mark for correctly distinguishing mol L⁻¹ (moles per litre) from g L⁻¹ (mass per litre); 1 mark for an explanation of when each is most useful; 1 mark for the conversion formula c (mol L⁻¹) = c (g L⁻¹) ÷ MM (with MM defined).

Sample response. The formula c = n ÷ V describes moles of solute per litre of solution — the total mixture of solute and solvent. When solute dissolves, it occupies space in the solution, so the final volume of solution is greater than the volume of solvent alone. Using only the solvent volume would overestimate the concentration. The correct volume is measured after the solute has fully dissolved and is typically read from a volumetric flask at the calibration mark.

Marking notes. 1 mark for explaining that adding solute increases the total volume beyond the solvent volume; 1 mark for stating that the V in c = n ÷ V is the volume of the final solution (accept reference to a volumetric flask or graduated container).

Sample response. MM(NaOH) = 22.99 + 16.00 + 1.008 = 39.998 g mol⁻¹ [1 mark]. n = m ÷ MM = 4.00 ÷ 39.998 = 0.1000 mol [1 mark]. V = 200 mL ÷ 1000 = 0.200 L [1 mark for explicit conversion]. c = n ÷ V = 0.1000 ÷ 0.200 = 0.500 mol L⁻¹ [1 mark].

Marking notes. 1 = correct molar mass; 1 = correct moles; 1 = mL converted to L shown; 1 = correct concentration with unit. Deduct no marks if rounding is applied consistently.

Sample response (a) — molar mass (2 marks). MM(CuSO₄·5H₂O) = 63.55 + 32.06 + 4(16.00) + 5[2(1.008) + 16.00] = 63.55 + 32.06 + 64.00 + 5(18.016) = 63.55 + 32.06 + 64.00 + 90.08 = 249.69 g mol⁻¹ [1 mark for correct structure; 1 mark for correct final value].

Sample response (b) — concentrations (4 marks).
Alpha: n = 24.96 ÷ 249.69 = 0.09996 mol; V = 400 ÷ 1000 = 0.400 L; c = 0.09996 ÷ 0.400 = 0.250 mol L⁻¹ [2 marks: 1 for correct working, 1 for correct answer].
Beta: n = 12.48 ÷ 249.69 = 0.04998 mol; V = 250 ÷ 1000 = 0.250 L; c = 0.04998 ÷ 0.250 = 0.200 mol L⁻¹ [2 marks].

Sample response (c) — comparison (2 marks). Alpha (0.250 mol L⁻¹) and Beta (0.200 mol L⁻¹) are not equivalent [1 mark]. Alpha is 25% more concentrated than Beta. The technician should either add more distilled water to Alpha to dilute it to 0.200 mol L⁻¹, or prepare a fresh batch for whichever bath is out of specification [1 mark for a correct course of action].

Sample response (a) — moles required (2 marks). n = c × V = 3.00 × 10⁻³ × 2.20 × 10⁶ = 6600 mol [1 mark for formula; 1 mark for correct answer].

Sample response (b) — volume of stock solution (2 marks). V = n ÷ c = 6600 ÷ 0.800 = 8250 L [1 mark for formula; 1 mark for correct answer].

Sample response (c) — restoring concentration (3 marks). Deficit concentration = 3.00 × 10⁻³ − 1.50 × 10⁻³ = 1.50 × 10⁻³ mol L⁻¹ [1 mark]. Additional moles needed: n = 1.50 × 10⁻³ × 2.20 × 10⁶ = 3300 mol [1 mark]. Volume of stock: V = 3300 ÷ 0.800 = 4125 L. Assumption: the pool volume remains constant at 2.20 × 10⁶ L (the volume added by the stock solution is negligible, or the pool is filled to the same level) [1 mark for assumption].

Sample response. Accurate volume measurement and unit conversion are critical in concentration calculations because the formula c = n ÷ V defines concentration as moles per litre. The most common error is substituting volumes in millilitres directly into the formula, producing a result that is 1000 times too small. For example, if a student calculates c = 0.50 ÷ 250 instead of c = 0.50 ÷ 0.250, they obtain 0.002 mol L⁻¹ rather than the correct 2.00 mol L⁻¹ — an error that would result in a preparation 1000 times more dilute than intended. In a hospital context, such as preparing an intravenous saline drip at a Sydney hospital, a concentration error of this magnitude could be life-threatening: a patient receiving a solution 1000 times more dilute than prescribed would receive almost no active solute, while a solution 1000 times more concentrated could cause severe electrolyte imbalance. A second important distinction is between mol L⁻¹ and g L⁻¹. Clinical labels on IV bags in Australian hospitals typically express concentration in g L⁻¹ (e.g. “9 g L⁻¹ NaCl” for normal saline), while the c = n ÷ V formula requires mol L⁻¹. Using a g L⁻¹ value directly in stoichiometric calculations without first converting via MM would give incorrect mole calculations; for example, treating 9 g L⁻¹ NaCl as 9 mol L⁻¹ would overstate the molarity by a factor of 58 (MM(NaCl) ≈ 58 g mol⁻¹). At the BHP mineral-processing plant, copper(II) sulfate baths must be maintained at precise mol L⁻¹ concentrations for consistent electroplating; expressing these in g L⁻¹ without converting would lead to incorrect dosing of reagents. Correct solution preparation technique is equally important. Dissolving a solid in a fixed volume of solvent (e.g. adding KNO₃ to 250 mL of water) differs from making up to 250 mL in a volumetric flask, because the dissolved solid increases the solution volume. A volumetric flask, with its precise calibration mark and narrow neck, removes operator judgement from this step: filling to the mark with distilled water and inverting to mix ensures the total volume is exactly 250 mL to within ±0.3 mL. This technique is essential in primary standard preparation for titrations used in analytical chemistry across Australian university and TAFE laboratories. In summary, unit conversion errors in c = n ÷ V introduce thousand-fold errors; confusion between mol L⁻¹ and g L⁻¹ invalidates stoichiometric results; and using a beaker instead of a volumetric flask for solution preparation introduces systematic volume errors. Rigorous attention to each of these steps is required for accurate chemical practice in both laboratory and industrial settings.

Marking criteria (9 marks). 1 = analyses the consequence of mL/L unit error (result 1000× too small, with numerical example). 1 = links unit error to a real-world consequence in a named Australian context (hospital, industry, etc.). 1 = correctly distinguishes mol L⁻¹ from g L⁻¹ and explains when each is used. 1 = gives a named Australian example illustrating the g L⁻¹ / mol L⁻¹ distinction. 1 = explains the V = solution (not solvent) error and its concentration consequence. 1 = describes correct volumetric flask technique (make-up-to-mark, invert, rinse). 1 = second named Australian context (distinct from the first). 1 = integrates all three error sources (unit conversion, unit type, preparation technique) into a coherent evaluation. 1 = reaches an explicit evaluative judgement about the relative severity/significance of the errors; uses precise chemical terminology throughout (mol L⁻¹, molar mass, volumetric flask, stoichiometry).