Free Atoms, molecules & stoichiometry questions for AS Level Chemistry (9701), part of Physical chemistry. Each question has the answer and the marking points an examiner would look for.
Generate a full Atoms, Molecules & Stoichiometry worksheet →All Physical chemistry questions20.0 cm3 of an acidified solution of iron(II) sulfate reacts exactly with 22.4 cm3 of 0.0200 mol dm−3 KMnO4.
MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
What is the concentration of Fe2+?
Answer: D
Which term is described by this statement?
“The amount of substance containing 6.02 × 1023 particles.”
Answer: C
20.0 cm3 of an acidified solution of iron(II) sulfate reacts exactly with 24.8 cm3 of 0.0200 mol dm−3 KMnO4.
MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
What is the concentration of Fe2+?
Answer: D
2.00 g of impure calcium carbonate was added to 50.0 cm3 of 2.00 mol dm−3 hydrochloric acid (an excess). The impurities do not react. The resulting solution was made up to 250 cm3. 25.0 cm3 portions of this solution needed 32.42 cm3 of 0.200 mol dm−3 sodium hydroxide for neutralisation.
Calculate the amount of HCl added to the calcium carbonate.
Calculate the amount of HCl left over after the reaction with calcium carbonate.
Calculate the amount of calcium carbonate in the sample.
Calculate the percentage purity of the calcium carbonate.
Suggest why a back titration is used rather than titrating calcium carbonate directly.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.100 mol 0.100 mol B1 | 1 |
| (b) | 0.0648 mol NaOH used = 6.48 × 10−3 mol per 25.0 cm3 C1 × 10 = 0.0648 mol A1 | 2 |
| (c) | 0.0176 mol HCl reacted = 0.0352 mol C1 CaCO₃ = ½ × this = 0.0176 mol A1 | 2 |
| (d) | 88.0% 88.0% B1 | 1 |
| (e) | CaCO₃ insoluble / slow reaction calcium carbonate is insoluble so its reaction with acid is slow and the end-point is unclear B1 | 1 |
A student finds the relative molecular mass of a volatile liquid, propanone, by vaporising 0.113 g of it in a gas syringe at 373 K and 1.01 × 105 Pa. The vapour occupies 60 cm3.
State two assumptions of the ideal gas model.
Calculate the relative molecular mass of the liquid.
The empirical formula of a second compound is CH₂O and its Mᵣ is 60. Deduce its molecular formula.
Calculate the volume of oxygen, at room conditions, needed to burn 0.0100 mol of hexane completely.
A 25.0 cm³ sample of 0.100 mol dm−3 sodium carbonate reacts with hydrochloric acid. Calculate the volume of 0.200 mol dm−3 HCl needed for complete reaction.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | negligible volume of molecules; no intermolecular forces molecules have negligible volume B1 no intermolecular forces (elastic collisions, random motion) B1 | 2 |
| (b) | 57.8 n = pV/RT with V in m3 M1 n = 1.96 × 10−3 mol M1 Mᵣ = 57.8 A1 | 3 |
| (c) | C₂H₄O₂ 60/30 = 2, so C₂H₄O₂ B1 | 1 |
| (d) | 2.28 dm³ C₆H₁₄ + 9½O₂ → 6CO₂ + 7H₂O M1 0.0950 mol O₂ M1 × 24.0 = 2.28 dm3 A1 | 3 |
| (e) | 25.0 cm³ Na₂CO₃ + 2HCl, moles HCl = 0.00500 M1 25.0 cm3 A1 | 2 |
A 1.50 g sample of limestone (impure CaCO3, with unreactive impurities) is added to 50.0 cm3 of 1.00 mol dm−3 HCl, which is in excess. When the reaction has finished, 25.0 cm3 of the remaining solution is titrated with 0.250 mol dm−3 NaOH. The mean titre is 40.06 cm3.
Calculate the amount of HCl present initially, and the amount remaining in 25.0 cm3 of the final solution.
Calculate the percentage by mass of CaCO3 in the limestone.
Explain why this back-titration method is used rather than titrating the limestone directly.
Calculate the volume of CO2 produced at room conditions.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.05 mol; 0.01 mol 1 × 0.050 B1 0.25 × 40.06 ÷ 1000 B1 | 2 |
| (b) | 100% HCl remaining in whole solution (× 2) C1 HCl reacted = initial − remaining C1 ÷ 2 and × Mr (100.1) C1 100 A1 | 4 |
| (c) | limestone is insoluble/reacts slowly so end-point cannot be seen directly; excess acid ensures complete reaction CaCO3 insoluble / slow reaction B1 complete reaction ensured by excess acid; excess measured accurately B1 | 2 |
| (d) | 360 cm3 mol CO2 = mol CaCO3 C1 360 A1 | 2 |
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