Chemistry • Year 11 • Module 2 • Lesson 20
Module 2 Review
Lock in the core vocabulary, formulas, and concepts from all three Inquiry Questions before tackling harder Module 2 problems.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: mole, molar mass, Avogadro’s number, molar volume, concentration, primary standard, titration, equivalence point, limiting reagent, percentage yield, percentage purity, gravimetric analysis, concordant titres, empirical formula. 14 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The SI unit of amount of substance; one of these contains 6.022 × 1023 elementary entities. | |
| 1.2 | The mass of one mole of a substance in g/mol; numerically equal to the sum of atomic masses times subscripts. | |
| 1.3 | The number of particles in one mole of any substance: 6.022 × 1023 mol−1. | |
| 1.4 | The volume occupied by one mole of any gas at a specified temperature and pressure; 22.71 L/mol at STP. | |
| 1.5 | The amount of solute (in mol) dissolved in one litre of solution; calculated by c = n ÷ V. | |
| 1.6 | A pure, stable substance of known composition used to prepare a solution of precisely known concentration. | |
| 1.7 | A volumetric technique where a solution of known concentration is added to an unknown solution until stoichiometric equivalence is reached. | |
| 1.8 | The point in a titration at which the moles of titrant added are stoichiometrically equivalent to the moles of analyte in the flask. | |
| 1.9 | The reactant that is completely consumed first, stopping the reaction; identified by comparing n ÷ coefficient. | |
| 1.10 | The ratio of actual yield to theoretical yield, expressed as a percentage; always ≤ 100% in a real experiment. | |
| 1.11 | The mass of pure substance as a percentage of the total sample mass; applied before stoichiometry. | |
| 1.12 | A quantitative technique that uses the mass of a precipitate to determine the concentration or amount of an analyte. | |
| 1.13 | Titration results that agree within 0.10 mL of each other; the rough titre is discarded before these are averaged. | |
| 1.14 | The simplest whole-number ratio of atoms in a compound, derived from percentage composition data. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 14 marks (1 T/F + 1 correction each)
2.1 The molar volume of any gas at STP (NESA standard: 0°C, 100 kPa) is 24.8 L/mol. T / F
2.2 When converting mL to L for concentration calculations, you divide by 1000. T / F
2.3 The limiting reagent is always the reactant with the smaller mass. T / F
2.4 Percentage purity is applied to a reactant sample before performing stoichiometry. T / F
2.5 In the dilution formula c1V1 = c2V2, V2 represents the volume of water added to the concentrated solution. T / F
2.6 NaOH solution can be used as a primary standard because it is easy to obtain and dissolves readily in water. T / F
2.7 In a titration back-calculation, the concentration of the unknown is calculated using the volume of the titrant (not the volume of the flask/aliquot). T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 10 marks (1 per blank)
Word bank:
balanced equation · coefficient · concordant · equivalence · limiting reagent · mole · molar mass · purity · ratio · theoretical yield
The central concept linking every calculation in Module 2 is the ___________, the SI unit of amount of substance. To convert between mass and moles, you divide the mass by the ___________ of the substance. In stoichiometry, the ___________ provides the mole ___________ that converts moles of one substance into moles of another. The reactant with the smaller n ÷ ___________ value is the ___________, and all products are calculated from that reactant alone. The maximum possible mass of product is the ___________. Before performing stoichiometry on an impure sample, the percentage ___________ must be applied first. In a titration, the ___________ point is reached when moles of titrant exactly match moles of analyte, and only ___________ titres (within 0.10 mL) are averaged to determine the final result.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the module. 8 marks (2 each)
4.1 What is the function of Avogadro’s number (NA) in mole calculations, and what is its value?
4.2 Why must the volume (V) be expressed in litres when using the formula c = n ÷ V?
4.3 In a titration, why is the rough titre excluded when calculating the average titre?
4.4 What is the role of a primary standard in preparing a standard solution, and why cannot NaOH be used as one?
5. Build a concept map — the mole as universal bridge
Draw labelled arrows between the six terms below to show how they connect across Module 2. Each arrow must carry a linking phrase (e.g. “converts to”, “uses ratio from”, “divides by”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: moles (n) · mass (m) · particles (N) · concentration (c) · gas volume (V) · balanced equation.
Q1 — Term–definition match
1.1 mole • 1.2 molar mass • 1.3 Avogadro’s number • 1.4 molar volume • 1.5 concentration • 1.6 primary standard • 1.7 titration • 1.8 equivalence point • 1.9 limiting reagent • 1.10 percentage yield • 1.11 percentage purity • 1.12 gravimetric analysis • 1.13 concordant titres • 1.14 empirical formula.
Q2 — True / false with correction
2.1 False. The NESA standard molar volume at STP (0°C, 100 kPa) is 22.71 L/mol. 24.8 L/mol is the value at RTP (25°C, 100 kPa).
2.2 True. Dividing mL by 1000 converts to L (e.g. 250 mL ÷ 1000 = 0.250 L).
2.3 False. The limiting reagent is identified by the smallest n ÷ coefficient value, not the smallest mass. A reagent with a smaller mass but smaller coefficient may still be in excess.
2.4 True. Percentage purity gives the mass of pure reactant available; this is always applied before stoichiometry begins.
2.5 False. In c1V1 = c2V2, V2 is the total final volume of the diluted solution, not just the volume of water added.
2.6 False. NaOH cannot be used as a primary standard because it is hygroscopic (absorbs water from the atmosphere) and may also absorb CO2, meaning its true concentration cannot be accurately determined by mass alone.
2.7 False. The concentration of the unknown is calculated using the volume of the flask (aliquot volume), not the titre volume. The titre is used to find n(standard), which is then used via ratio to find n(unknown), and c(unknown) = n(unknown) ÷ V(flask).
Q3 — Cloze paragraph
In order: mole / molar mass / balanced equation / ratio / coefficient / limiting reagent / theoretical yield / purity / equivalence / concordant.
Q4.1 — Function of Avogadro’s number
Avogadro’s number (NA = 6.022 × 1023 mol−1) converts between the number of individual particles (atoms, molecules, ions) and the amount in moles. It is used in the formula N = n × NA, allowing chemists to count particles indirectly by measuring macroscopic amounts.
Q4.2 — Why V must be in litres for c = n ÷ V
Concentration is defined in mol/L (mol per litre), so the formula c = n ÷ V requires V in litres to give the correct units. If V is given in mL, it must first be divided by 1000. Using mL directly would give a concentration 1000 times too small.
Q4.3 — Why rough titre is excluded
The rough titre is performed to estimate the endpoint and is conducted quickly, so the endpoint may be overshot; it is deliberately imprecise and unreliable. Only the subsequent careful titrations that agree within 0.10 mL (concordant titres) are averaged to minimise random error.
Q4.4 — Role of primary standard and why NaOH fails
A primary standard is a substance of precisely known purity, high molar mass, and stability that can be accurately weighed to prepare a solution of precisely known concentration. NaOH fails because it is hygroscopic (absorbs atmospheric water) and reacts with CO2, so the actual mass of pure NaOH present is unknown; this makes it impossible to calculate an accurate concentration by mass.
Q5 — Sample concept map
Award 1 mark per valid labelled arrow. Correct arrows include:
- mass — divided by molar mass gives → moles
- moles — multiplied by NA gives → particles
- moles — divided by volume (L) gives → concentration
- moles — multiplied by molar volume gives → gas volume
- balanced equation — provides mole ratio to convert → moles
- concentration — multiplied by volume (L) gives → moles
Award 1 mark per correctly labelled, directionally accurate arrow (minimum 6 required).