This quiz covers Lessons 1–5: Classification of Matter, Properties, Filtration, Crystallisation, Distillation, Chromatography, and Gravimetric Analysis. 20 multiple choice + 3 short answer questions.
Lesson Summaries — Quick Review
Matter is classified as pure substances (elements and compounds) or mixtures (homogeneous and heterogeneous). The classification is based on particle-level composition — pure substances have uniform, fixed composition; mixtures have variable composition. Elements contain only one type of atom; compounds have two or more elements chemically bonded in fixed ratios.
Pure substances have sharp, fixed melting and boiling points; mixtures melt and boil over a range. Compounds have completely different properties from their constituent elements — new bonds create new structures with new properties. Physical properties (MP, BP, conductivity, solubility, density) are used to identify and classify unknown substances.
Filtration separates insoluble solids from liquids by particle size — solid stays as residue; liquid passes as filtrate. Crystallisation separates dissolved solids from solution by exploiting decreasing solubility on cooling — dissolved solid crystallises out. Slow cooling gives larger, purer crystals. The techniques are often used in sequence.
Simple distillation separates liquids with large BP differences; fractional distillation uses a fractionating column for close BPs. Chromatography separates dissolved compounds by differential affinity for stationary/mobile phases. Rf = distance moved by component ÷ distance moved by solvent front. Rf values identify unknown substances by comparison with standards under identical conditions.
Gravimetric analysis determines the mass of a dissolved ion by precipitating it as an insoluble solid, filtering, drying to constant mass, weighing, and applying stoichiometry. Excess reagent ensures complete precipitation. Always dry to constant mass. Choosing the right technique requires identifying the key physical property that differs between components.
Score updates as you answer each question. Short answer marks are self-assessed.
1. Which of the following is a compound? L01
2. Bronze is an alloy of copper and tin. How should it be classified? L01
3. Which correctly distinguishes an element from a compound? L01
4. A student examines a clear liquid under a microscope and cannot distinguish different components. What can they conclude? L01 HOT
5. A substance melts over the range 48–72°C. What does this indicate? L02
6. Solid sodium chloride (NaCl) does not conduct electricity, but when melted or dissolved in water it does. The best explanation is: L02
7. Which data table row is consistent with an ionic compound? L02 Data
| Option | MP | Conductivity (solid) | Conductivity (dissolved) |
|---|---|---|---|
| A | −6°C (sharp) | None | None |
| B | 1538°C (sharp) | Excellent | — |
| C | 40–70°C (range) | None | Low |
| D | 801°C (sharp) | None | Excellent |
8. During filtration of a mixture of sand and salt water, where does the NaCl end up? L03
9. Why does slow cooling during crystallisation produce purer crystals than rapid cooling? L03
10. A student filters a copper sulfate solution hoping to collect CuSO₄ crystals. They collect nothing on the filter paper. The most likely reason is: L03 HOT
11. The solubility of potassium nitrate (KNO₃) at 60°C is 110 g per 100 mL water, and at 20°C is 31 g per 100 mL water. If a saturated solution is prepared at 60°C and cooled to 20°C, which correctly describes the mass of KNO₃ that remains in solution in 100 mL? L03 Data
12. A fractionating column is used in fractional distillation because it: L04
13. On a chromatography strip, a spot moves 4.8 cm and the solvent front moves 8.0 cm. The Rf value is: L04
14. A component with a high Rf value (close to 1) has: L04
15. The Rf value of a compound measured using hexane solvent is 0.72. A reference from a different lab using ethanol solvent also shows Rf = 0.72 for a known compound. What is the most valid conclusion? L04 HOT
16. In gravimetric analysis, why must the precipitate be dried to constant mass before weighing? L05
17. In a gravimetric analysis, 0.233 g of BaSO₄ precipitate is collected. The molar mass of BaSO₄ is 233.4 g mol⁻¹ and SO₄²⁻ is 96.1 g mol⁻¹. The mass of SO₄²⁻ in the sample is approximately: L05
18. A mixture of ethanol (BP 78°C) and water (BP 100°C) needs to be separated. Which technique is most appropriate? L04/L05
19. A food manufacturer needs to confirm the exact concentration of calcium chloride (CaCl₂) added to a batch of processed food. Which technique is most appropriate? L05
20. A researcher has a solution containing three different dissolved dyes and needs to: (a) determine how many components are present, and (b) identify each one by comparison with known standards. Which technique best addresses both needs simultaneously? L05 HOT
Attempt all three questions before checking answers.
21. A sample of seawater contains dissolved NaCl and fine sand particles. Outline a two-step procedure to obtain both dry sand and dry NaCl crystals from this sample. For each step, name the technique and state the property it exploits. 4 MARKS
22. On a paper chromatography strip, two spots are observed. Spot A is at 2.4 cm and Spot B is at 6.0 cm from the origin. The solvent front is at 8.0 cm. Reference data shows: Compound P has Rf = 0.30, Compound Q has Rf = 0.70, Compound R has Rf = 0.75. (a) Calculate the Rf values for Spots A and B. (b) Identify each spot if possible, or explain why identification is uncertain. 4 MARKS
23. A student performs gravimetric analysis to determine the mass of barium ions (Ba²⁺) in a sample. They add excess sodium sulfate solution and collect 0.699 g of dry BaSO₄ precipitate. (a) Write the ionic equation for the precipitation reaction. (b) Calculate the mass of Ba²⁺ in the sample. Show full working. [M(BaSO₄) = 233.4 g mol⁻¹; M(Ba) = 137.3 g mol⁻¹] 5 MARKS
26. (L01 — 3 marks) Classify each of the following as an element, compound, homogeneous mixture, or heterogeneous mixture. Justify each classification: (a) Carbon dioxide gas (CO₂), (b) A sample of air, (c) A glass of orange juice with pulp visible. 3 MARKS
27. (L01 — 2 marks) Explain the difference between a physical change and a chemical change. Give one example of each that involves water. 2 MARKS
28. (L02 — 3 marks) A student measures the following for two samples: Sample A — melting point 800°C, conducts electricity when dissolved in water, does not conduct as a solid. Sample B — melting point −117°C, does not conduct electricity in any state. Use these properties to identify the likely bonding type of each sample, and suggest one possible substance for each. 3 MARKS
29. (L02 — 3 marks) Explain why a pure substance has a sharp, fixed melting point while a mixture melts over a temperature range. Use particle-level reasoning. 3 MARKS
30. (L03 — 3 marks) A student is attempting to purify rock salt (a mixture of NaCl and sand) to obtain pure dry NaCl crystals. Describe a complete procedure including: the technique used to remove sand, the technique used to recover dry salt crystals, and one precaution for each step. 3 MARKS
31. (L03 — 3 marks) A student dissolves 8.5 g of KNO₃ in 20 mL of hot water, then cools the solution. Explain what happens at the particle level during crystallisation. Why does slow cooling produce larger, purer crystals than rapid cooling? 3 MARKS
32. (L04 — 4 marks) A wine has an ethanol content of 13% by volume. A chemist wants to produce a concentrated ethanol solution using laboratory distillation. (a) Explain the principle behind simple distillation for separating ethanol from water. (b) Why would fractional distillation be more effective than simple distillation for this separation? (c) What piece of equipment in fractional distillation provides the additional separation? 4 MARKS
33. (L04 — 3 marks) A student runs a paper chromatography experiment on a food dye and obtains the following results: solvent front travels 9.6 cm; dye spot 1 travels 7.2 cm; dye spot 2 travels 3.6 cm. (a) Calculate the Rf value for each dye spot. (b) The student re-runs the experiment with twice the volume of solvent. Predict whether the Rf values will change. Explain. 3 MARKS
34. (L05 — 4 marks) A chemist uses gravimetric analysis to determine the mass of chloride ions in a 50.00 mL sample of tap water. Excess silver nitrate (AgNO₃) is added to precipitate all Cl⁻ as AgCl. The dried precipitate has a mass of 0.287 g. (a) Write a word equation for the precipitation reaction. (b) Explain why excess AgNO₃ is used. (c) The molar mass of AgCl = 143.4 g/mol and Cl = 35.5 g/mol. Calculate the mass of Cl⁻ in the original sample. (d) Express this as a concentration in g/L. 4 MARKS
35. (L05 — 3 marks) A food scientist needs to separate and identify the individual colourings in a green sports drink that contains a mixture of blue and yellow dyes. The dyes are all water-soluble and non-volatile. Recommend the most appropriate separation technique and justify your choice over two alternatives (distillation and filtration). 3 MARKS
1. C CO₂ is a compound (C and O bonded). Ar is an element; iron+sulfur is a mixture; air is a mixture.
2. A Alloys are homogeneous mixtures — metals are physically combined without fixed molar ratios or chemical bonds between all components.
3. D The only chemically precise distinction: one type of atom (element) vs two or more types chemically bonded in fixed ratios (compound).
4. B Optical microscopy cannot distinguish dissolved ions from pure liquid molecules. Melting point testing is needed to confirm purity.
5. C A 24°C melting range confirms mixture. Pure substances — even complex ones — melt at a single temperature.
6. A Ions are immobile in the solid lattice. Melting or dissolving frees them to move and carry charge.
7. D D shows: sharp MP (801°C) = pure substance; no solid conductivity + excellent dissolved conductivity = ionic compound. A is covalent molecular; B is metal; C is a mixture.
8. B Dissolved NaCl passes through with the filtrate. Only insoluble solids are retained as residue.
9. C Slow cooling → ordered lattice → impurity exclusion. Rapid cooling → impurities trapped.
10. A CuSO₄ is dissolved. Dissolved substances pass through filter paper — filtration cannot separate them.
11. D At 20°C, maximum solubility = 31 g per 100 mL. The rest (110 − 31 = 79 g) crystallises out. So 31 g remains dissolved.
12. B The fractionating column creates multiple condensation/vaporisation cycles, enriching the vapour in the more volatile component at the top.
13. C Rf = 4.8 ÷ 8.0 = 0.60
14. D High Rf → moved far → carried strongly by mobile phase → high affinity for mobile phase, low for stationary phase.
15. A Rf values are only comparable under identical conditions. Different solvents give different Rf values for the same compound.
16. B Residual moisture adds to the measured mass → overestimates the result. Dry to constant mass to ensure all water is removed.
17. C n(BaSO₄) = 0.233 ÷ 233.4 = 0.000999 mol. n(SO₄²⁻) = 0.000999 mol. m(SO₄²⁻) = 0.000999 × 96.1 = 0.0960 ≈ 0.0961 g.
18. D 22°C BP difference requires fractional distillation — a fractionating column provides multiple separation cycles.
19. A Gravimetric analysis gives a quantitative mass measurement of a specific ion — the only technique listed that does this.
20. C Chromatography simultaneously shows the number of components (number of spots) and allows identification by Rf comparison to standards.
Q26 (3 marks): (a) CO₂: compound — composed of two different elements (C and O) chemically bonded in a fixed ratio (1:2). It is a pure substance with a fixed composition. Cannot be separated by physical means. (b) Air: homogeneous mixture — contains N₂ (~78%), O₂ (~21%), Ar, CO₂ etc., all gases mixed at the molecular level. Uniform composition throughout; can be separated by physical means (fractional distillation of liquid air). (c) Orange juice with pulp: heterogeneous mixture — the pulp (solid fibres) is visibly distinct from the liquid juice. Non-uniform composition; the pulp could be removed by filtration, leaving "smooth" orange juice.
Q27 (2 marks): Physical change: the chemical identity (formula) of the substance is preserved; no new substance is formed; generally reversible. Example: water freezing to ice — both are H₂O, just different states; melting reverses the change. Chemical change: new substances with different chemical formulas and properties are formed; generally not easily reversible. Example: electrolysis of water (2H₂O → 2H₂ + O₂) — the products H₂ and O₂ are completely different substances from water; the process requires energy input and is not spontaneously reversible.
Q28 (3 marks): Sample A: Ionic compound — high MP (800°C requires breaking strong electrostatic forces in ionic lattice), conducts when dissolved in water (ions dissociate into solution and become mobile charge carriers), insulates as solid (ions fixed in rigid lattice, cannot move). Possible substance: NaCl (MP 801°C). Sample B: Covalent molecular substance — very low MP (−117°C indicates only weak intermolecular forces between small molecules), non-conducting in all states (no ions and no delocalised electrons). Possible substance: ethanol C₂H₅OH (MP exactly −117°C).
Q29 (3 marks): Pure substance: all particles are chemically identical — every particle-particle interaction is the same strength. At the melting point, all particles simultaneously acquire enough thermal energy to overcome these identical forces, converting entirely from solid to liquid at one fixed temperature — a sharp melting point is observed. Mixture: different types of particles interact with each other in various ways. Some particle-particle interactions are weaker than others (e.g. between different molecule types). The weakest interactions are overcome first at lower temperatures — the substance begins to soften. As temperature increases further, progressively stronger interactions are overcome — melting is gradual over a temperature range. A pure substance can therefore be identified by its sharp, reproducible melting point, while a mixture has a depressed and broadened melting range.
Q30 (3 marks): Step 1 — Filtration to remove sand: dissolve rock salt in hot distilled water (NaCl dissolves, sand does not). Set up a filter funnel with filter paper in a conical flask. Pour the mixture through — NaCl solution (filtrate) passes through; sand (residue) is retained. Precaution: use excess hot water to ensure all NaCl dissolves; rinse the filter paper with a small amount of distilled water to collect remaining NaCl. Step 2 — Crystallisation to recover dry NaCl: pour the filtrate into an evaporating dish. Heat gently until the volume reduces by ~half and a crystal crust begins to form around the edge. Remove from heat and allow to cool slowly — NaCl crystallises. Filter to collect crystals. Dry in an oven at 100°C. Precaution: do not evaporate to complete dryness while on the hotplate — NaCl may spit from the dish; stop heating when the first crystals appear.
Q31 (3 marks): At high temperature, KNO₃ has high solubility — all ions remain in solution. As the solution cools, solubility decreases. When the concentration of dissolved ions exceeds the new (lower) solubility, the solution becomes supersaturated. K⁺ and NO₃⁻ ions begin to collide and arrange into the ionic crystal lattice — the lattice energy released (electrostatic attraction) makes crystallisation thermodynamically favourable. Slow cooling: gives ions time to migrate to the correct lattice positions, growing a well-ordered crystal structure. Few impurity ions are incorporated because the crystal grows slowly and selectively. Result: fewer but larger, purer crystals. Rapid cooling: many nucleation sites form simultaneously, all competing for available ions. Crystals grow quickly — ions don't have time to position perfectly, impurities become trapped. Result: many small, impure crystals.
Q32 (4 marks): (a) Simple distillation works by exploiting the difference in boiling points. Ethanol (BP 78°C) is more volatile than water (BP 100°C). The mixture is heated — ethanol-enriched vapour rises first into the condenser, cools, and is collected as liquid. Water preferentially remains in the flask. The separation is driven by the difference in vapour pressure at any given temperature. (b) Because ethanol and water are fully miscible with only a 22°C BP difference, their vapour compositions at any given temperature contain both components. A single simple distillation gives only partially purified ethanol (the vapour still contains ~50% water at typical distillation temperatures). Fractional distillation performs many repeated vaporisation-condensation cycles as vapour rises up the column, progressively concentrating the more volatile component (ethanol) in the vapour and the less volatile (water) as condensate running back into the flask — achieving much higher purity in a single run. (c) The fractionating column — packed with glass beads, rings, or fitted with plates — provides large surface area for repeated vaporisation-condensation equilibria along the length of the column.
Q33 (3 marks): (a) Rf(spot 1) = 7.2 / 9.6 = 0.75. Rf(spot 2) = 3.6 / 9.6 = 0.375 (round to 0.38). (b) Rf values will not change. Rf is defined as the ratio of the distance travelled by the spot to the distance travelled by the solvent front. Using twice the volume of solvent allows the solvent front to travel further — but the spot will also travel proportionally further. Because both distances scale together, the ratio remains constant. Rf is a characteristic property of a substance under defined conditions (same solvent, stationary phase, temperature) — independent of solvent volume or paper length.
Q34 (4 marks): (a) Silver nitrate + tap water chloride → silver chloride (white precipitate) + nitrate ions remaining in solution. (Ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).) (b) Excess AgNO₃ ensures all Cl⁻ ions have reacted and been precipitated — if AgNO₃ were limiting, some Cl⁻ would remain in solution and not be captured in the precipitate, causing an underestimate of Cl⁻ mass. Excess drives the equilibrium strongly toward product (Le Chatelier). (c) Moles AgCl = 0.287 g ÷ 143.4 g/mol = 0.002001 mol. Since Ag⁺:Cl⁻ = 1:1, moles Cl⁻ = 0.002001 mol. Mass Cl⁻ = 0.002001 × 35.5 = 0.0710 g. (d) Volume = 50.00 mL = 0.05000 L. Concentration = 0.0710 g ÷ 0.05000 L = 1.42 g/L.
Q35 (3 marks): Most appropriate technique: paper chromatography (or thin-layer chromatography). Justification: the dyes are all water-soluble, non-volatile, and present together in solution — they need to be separated based on their individual chemical properties, not physical state or boiling point. Chromatography exploits each dye's different affinity for the stationary phase (paper cellulose) vs the mobile phase (water solvent) — each dye travels a characteristic distance, producing distinct spots that can be identified by Rf values and compared to known standards. Against distillation: the dyes are non-volatile — they will not vaporise and pass into the condenser. Distillation would only separate the water solvent from all the dyes together, not separate the individual dyes from one another. Against filtration: all dyes are dissolved in solution — they are not solid particles. Filtration separates insoluble solids from liquids; a dissolved dye passes straight through filter paper with the solvent.
Q21 (4 marks): Step 1 — Filtration: pour the seawater through filter paper in a funnel. The technique exploits particle size — sand is insoluble and too large to pass through the filter paper; it is retained as residue. Wash with distilled water and dry the residue to obtain dry sand. (2 marks — technique + property + what is collected). Step 2 — Crystallisation: heat the filtrate (NaCl solution) in an evaporating basin to concentrate it, cool slowly to allow NaCl to crystallise, filter off the crystals, and dry. The technique exploits the decrease in solubility of NaCl with temperature. (2 marks — technique + property + procedure).
Q22 (4 marks): Rf(A) = 2.4 ÷ 8.0 = 0.30 (1 mark). Rf(B) = 6.0 ÷ 8.0 = 0.75 (1 mark). Spot A (Rf 0.30) = Compound P (Rf 0.30) — exact match (1 mark). Spot B (Rf 0.75) — does not exactly match Compound Q (0.70) or Compound R (0.75). Rf(B) = 0.75 matches Compound R exactly. However, if the Rf values of Q and R are close (0.70 vs 0.75), there may be some uncertainty due to measurement precision — the student should note that Spot B most likely matches Compound R but confirmation with a reference run would be advisable (1 mark for identification + noting Q/R uncertainty).
Q23 (5 marks): (a) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) (1 mark). (b) n(BaSO₄) = 0.699 ÷ 233.4 = 0.002996 mol (1 mark). From balanced equation: n(Ba²⁺) = n(BaSO₄) = 0.002996 mol [1:1 ratio] (1 mark). m(Ba²⁺) = n × M = 0.002996 × 137.3 = 0.411 g (1 mark). The sample contains 0.411 g of Ba²⁺ ions (1 mark for correct answer with units).
Tick when you have checked all your answers and noted areas to review.