| Exam date | Questions | Time | Maximum marks |
|---|---|---|---|
| 23 November 2025 | 60 | 120 minutes | 216 |
Final answers · Code 31
Match this code to your question paper. For Questions 49–60, choose the complete listed set; a single listed option is possible even in the multiple-correct section.
| Question | Answer | Question | Answer | Question | Answer |
|---|---|---|---|---|---|
| 1 | C | 21 | B | 41 | C |
| 2 | D | 22 | B | 42 | A |
| 3 | DROPPED | 23 | D | 43 | B |
| 4 | A | 24 | B | 44 | A |
| 5 | D | 25 | D | 45 | A |
| 6 | C | 26 | C | 46 | A |
| 7 | D | 27 | A | 47 | B |
| 8 | B | 28 | C | 48 | C |
| 9 | A | 29 | D | 49 | A |
| 10 | D | 30 | D | 50 | A |
| 11 | A | 31 | C | 51 | A, B, D |
| 12 | C | 32 | C | 52 | A, B, C, D |
| 13 | C | 33 | D | 53 | B, C, D |
| 14 | D | 34 | A | 54 | A, B, C, D |
| 15 | C | 35 | B | 55 | B, D |
| 16 | C | 36 | A | 56 | A, B, C, D |
| 17 | B | 37 | A | 57 | B, C |
| 18 | C | 38 | B | 58 | A, C |
| 19 | C | 39 | D | 59 | B, C, D |
| 20 | B | 40 | B | 60 | A, C, D |
Six worked explanations
Keep the official Code 31 paper open alongside these original explanations. The complete organiser solution PDF is linked above; the six expanded notes below explain the numerical reasoning step by step.
Question 34: Back titration: find the impurity
Answer AFirst find the acid added: 0.060 × 0.4 = 0.024 mol HCl. The excess acid consumes 0.020 × 0.4 = 0.008 mol NaOH, so 0.016 mol HCl reacted with the impurity. Since M(OH)₂ consumes two moles of HCl per mole, the impurity is 0.008 mol. Its mass is 0.008 × 65 = 0.520 g. The mass percentage is 100 × 0.520/30 = 1.73%, matching A. The chloride salt is not counted as consuming acid in the stated experiment.
Question 40: Kinetics: recognise repeated tenfold decreases
Answer BThe elementary unimolecular reaction is first order. Its concentration falls from 0.35 M to 0.035 M in 10 minutes, a factor of ten. Reaching 0.00035 M is a factor of 1000 below the original concentration, so three equal tenfold intervals are needed: t = 30 minutes. If calculated explicitly, k = ln(10)/10 = 0.2303 min⁻¹. The time data are in minutes, so the rate-constant unit here is min⁻¹.
Question 41: A weak acid reveals the pH of its buffer
Answer CThe original pH of 3.0 gives [H⁺] = 10⁻³ M. For the 0.20 M acid, Ka = x²/(C − x) = 10⁻⁶/0.199 ≈ 5.03 × 10⁻⁶; pKa ≈ 5.30. The mixture contains 25.0 mL × 0.20 M = 0.00500 mol propanoate and 25.0 mL × 0.10 M = 0.00250 mol acid. Their ratio is 2. Thus pH ≈ pKa + log 2 ≈ 5.30 + 0.30 = 5.60. The common final volume cancels in the ratio.
Question 42: Precipitation: compare threshold metal concentrations
Answer AA precipitate begins when its ion product reaches Ksp. With each anion at 0.010 M, MSO₃ needs [M²⁺] = 7 × 10⁻⁷/0.010 = 7 × 10⁻⁵ M. MF₂ needs [M²⁺] = 5 × 10⁻⁹/(0.010)² = 5 × 10⁻⁵ M. For M₃(PO₄)₂, [M²⁺]³(0.010)² = 10⁻²⁵, giving [M²⁺] = 10⁻⁷ M. The smallest threshold is reached first: phosphate, fluoride, then sulfite. Comparing Ksp values alone would ignore stoichiometry.
Question 43: Freezing point meets coordination chemistry
Answer BUse ΔTf = iKfm. The supplied values give i = 0.558/(1.86 × 0.100) = 3. Under the stated complete-ionisation assumption, each formula unit therefore gives three dissolved ions. [Cr(NH₃)₅Cl]Cl₂ dissociates into one complex cation and two chloride ions. The chloride inside the coordination sphere remains bound. This gives three particles and six donor atoms around chromium, matching option B.
Question 44: Aspirin: identify the limiting reagent before yield
Answer AThe two reactants are used in a 1:1 mole ratio. Salicylic acid supplies 155/138 ≈ 1.123 mol, while acetic anhydride supplies 105/102 ≈ 1.029 mol. Acetic anhydride is limiting. At 80% conversion, aspirin formed = 0.80 × 1.029 ≈ 0.824 mol. Its mass is 0.824 × 180 ≈ 148 g. Apply the conversion to the limiting amount, not to the combined starting mass.
Make the review count
For each missed problem, write the governing chemical idea before repeating the calculation. Check units, charges, reaction stoichiometry and which assumptions the question permits. In multiple-correct questions, justify every option independently.
Keep a short error log: question number, cause of error, corrected reasoning and a date to try it again. A solved paper becomes useful when you can reproduce the reasoning later.
Sources & verification
Exam information checked 20 October 2026. Subsequent official notices take precedence.