IGCSE Chemistry 0620 Paper 6 Alternative to Practical Practice Paper

Level: Year 10–11 (IGCSE) | Subject: Chemistry 0620 | Time: 60 minutes | Total: 40 marks | Parts: 26

Paper at a glance

Paper	Time	Marks	Weighting
Paper 6 (Alternative to Practical)	1 h	40	20%

Questions

Question 1 (9 marks) — Titration: sodium hydroxide and dilute sulfuric acid

(a) Name the apparatus used to measure exactly 25.0 cm³ of the acid, and state why a measuring cylinder is not suitable. (1 marks)

(b) Give two precautions to take when filling the burette, explaining each. (2 marks)

(c) Burette readings (cm³): 1st 0.00 to 24.90; 2nd 0.00 to 24.70; 3rd 24.70 to 49.45; 4th 0.50 to 25.20. Calculate each titre and the mean titre to be used (2 d.p.). (3 marks)

(d) The sodium hydroxide is 0.100 mol/dm³ and the reaction is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the moles of sodium hydroxide used and the concentration of the acid (mol/dm³, 3 s.f.). (2 marks)

(e) Phenolphthalein is used as the indicator. State the colour change at the end-point. (1 marks)

Question 2 (8 marks) — Preparing copper(II) sulfate crystals from copper(II) oxide

(a) Excess copper(II) oxide is added, stirred and then filtered off. Explain (i) why the oxide is added in excess (ii) why the mixture is filtered. (2 marks)

(b) Name the two stages used to obtain crystals from the filtrate. (2 marks)

(c) 25.0 cm³ of 0.500 mol/dm³ sulfuric acid is used and Mr of CuSO₄·5H₂O = 250. Calculate the theoretical yield (g, 3 s.f.). (2 marks)

(d) The mass of crystals actually obtained is 2.60 g. Calculate the percentage yield. (1 marks)

(e) Why are the crystals washed with a little cold water rather than hot water? (1 marks)

Question 3 (9 marks) — The rate of reaction between magnesium and dilute hydrochloric acid

(a) Name the apparatus used to measure accurately the volume of hydrogen collected. (1 marks)

(b) Each run uses the same mass and surface area of magnesium ribbon at a constant temperature. State the purpose of this. (1 marks)

(c) Data: time 0/15/30/45/60/75/90/105/120/135 s, volume of hydrogen 0/14/26/36/44/50/54/57/58/58 cm³. Calculate the average rate from 0 to 15 s (cm³/s, 2 s.f.) and state what this value represents. (2 marks)

(d) Calculate the average rate between 30 s and 60 s and explain why it is lower than the initial rate. (2 marks)

(e) After 120 s the volume stays at 58 cm³. Explain why, and calculate the moles of hydrogen formed (molar volume 24 dm³ mol⁻¹, 2 s.f.). (2 marks)

(f) If the same mass of powdered magnesium were used, how would the curve differ? (1 marks)

Question 4 (7 marks) — Chromatography and Rf values

(a) Why is the baseline drawn in pencil rather than with a pen? (1 marks)

(b) Why must the level of the solvent stay below the spots? (1 marks)

(c) The solvent front is 7.2 cm from the baseline; spot A is 2.16 cm and spot C is 6.12 cm from the baseline. Calculate the Rf values of A and C (2 s.f.). (2 marks)

(d) A third spot B is 3.60 cm from the baseline. Calculate its Rf value and describe how to decide whether B is the same substance as a known dye. (2 marks)

(e) Why compare Rf values rather than the distances the spots move? (1 marks)

Question 5 (7 marks) — Water of crystallisation and heating to constant mass

(a) Name the apparatus used to heat and to weigh the solid, and state why it is heated to constant mass. (1 marks)

(b) The sample weighs 5.00 g before heating and 3.20 g at constant mass. Calculate the mass of water lost and its percentage of the original sample (3 s.f.). (2 marks)

(c) Mr of the anhydrous salt is 160 and Mr of water is 18. Calculate how many moles of water combine with each mole of anhydrous salt (x, 2 s.f.). (2 marks)

(d) The anhydrous salt is CuSO₄. Write the formula of the hydrated salt. (1 marks)

(e) If a little of the white solid spattered out during heating, how would the calculated value of x be affected? (1 marks)

Answers and marking notes

Question 1 — Titration: sodium hydroxide and dilute sulfuric acid

(a) A 25.0 cm³ volumetric pipette (1). A measuring cylinder has coarse graduations (about ±0.5 cm³) and is not accurate enough for a titration. [1 marks]

(b) Rinse the burette with a little of the solution so residual water does not dilute it; remove the air bubble from the jet, otherwise the reading is too large; read with the eye level with the bottom of the meniscus to avoid parallax. Any two, 1 mark each. [2 marks]

(c) Titres are 24.90, 24.70, 24.75 and 24.70 cm³ (1); the first is a rough titration and is not used (1); the other three agree within 0.10 cm³, mean = (24.70 + 24.75 + 24.70) / 3 = 24.72 cm³ (1, must be 2 d.p.). [3 marks]

(d) n(NaOH) = 0.100 × 24.72 / 1000 = 2.47 × 10⁻³ mol (1); n(H₂SO₄) = 2.47 × 10⁻³ / 2, c = 1.24 × 10⁻³ / 0.0250 = 0.0494 mol/dm³ (1). Omitting the 1:2 ratio gives 0.0989 mol/dm³ and scores no second mark. [2 marks]

(e) Colourless to pale pink, the pink lasting about 30 seconds (1). The acid is in the flask and the alkali in the burette, so the change is colourless to pink. [1 marks]

Question 2 — Preparing copper(II) sulfate crystals from copper(II) oxide

(a) (i) To ensure all the acid reacts and no acid is left, so the crystals are pure (1); (ii) to remove the unreacted excess black copper(II) oxide solid (1). [2 marks]

(b) Heat to evaporate and concentrate until crystals just begin to appear (1); then cool to crystallise, filter, wash with a little cold water and dry (1). [2 marks]

(c) n(H₂SO₄) = 0.500 × 0.0250 = 0.0125 mol (1); the ratio is 1:1, so m = 0.0125 × 250 = 3.13 g (1; accept 3.125 g). [2 marks]

(d) 2.60 / 3.125 × 100 = 83.2% (1; 83.1% from 3.13 g is also accepted). [1 marks]

(e) The crystals are far more soluble in hot water, so hot water would dissolve them and reduce the yield (1). [1 marks]

Question 3 — The rate of reaction between magnesium and dilute hydrochloric acid

(a) A gas syringe (1); an inverted measuring cylinder over a water trough is also accepted. [1 marks]

(b) It controls the variables (a fair test) so that the volume change is caused only by concentration and time, making the runs comparable (1). [1 marks]

(c) (14 – 0) / 15 = 0.93 cm³/s (1); it represents the initial rate of reaction (1). [2 marks]

(d) (44 – 26) / 30 = 0.60 cm³/s (1); the acid concentration falls as it is used up, so collisions become less frequent and the rate decreases (1). [2 marks]

(e) The reaction has stopped because the limiting reactant, magnesium, has been used up (1); n(H₂) = 58 / 24000 = 2.4 × 10⁻³ mol (1). [2 marks]

(f) The curve would be steeper (a larger initial gradient, a faster reaction) but the final volume collected would be the same (1). [1 marks]

Question 4 — Chromatography and Rf values

(a) The dyes in pen ink are soluble in the solvent and would travel up with it, contaminating the chromatogram (1). [1 marks]

(b) If the solvent were above the spots, the spots would dissolve into it and no separation would occur (1). [1 marks]

(c) Rf(A) = 2.16 / 7.2 = 0.30 (1); Rf(C) = 6.12 / 7.2 = 0.85 (1). [2 marks]

(d) Rf(B) = 3.60 / 7.2 = 0.50 (1); run the known dye alongside B with the same solvent, paper and temperature, and only an identical Rf value identifies the same substance (1). Comparing colour intensity alone scores nothing. [2 marks]

(e) Rf divides out the distance moved by the solvent front, so results from different paper sizes or running times can still be compared (1). [1 marks]

Question 5 — Water of crystallisation and heating to constant mass

(a) A crucible (or evaporating basin) with a Bunsen burner, and a balance; heating is repeated until two consecutive masses differ by no more than 0.01 g, so all the water of crystallisation has been removed (1). [1 marks]

(b) 5.00 – 3.20 = 1.80 g (1); 1.80 / 5.00 × 100 = 36.0% (1). [2 marks]

(c) n(H₂O) = 1.80 / 18 = 0.100 mol (1); n(anhydrous salt) = 3.20 / 160 = 0.0200 mol, x = 0.100 / 0.0200 = 5.00 (1). [2 marks]

(d) CuSO₄·5H₂O (1). [1 marks]

(e) The final mass would be too low, so the calculated mass of water, and therefore x, would be too large (1). [1 marks]

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