IGCSE Chemistry 0620/62 May/June 2025 Past Paper with Answers

Question 1

Clean magnesium ribbon is heated with steam and the hydrogen gas produced is collected in a piece of apparatus that can both collect and measure the volume of a gas. What is the most suitable name for this apparatus?

1. a gas syringe
2. a measuring cylinder
3. a conical flask
4. a filter funnel

Question 2

Magnesium ribbon becomes coated with a dull layer of magnesium oxide when left in air. How should the magnesium ribbon be cleaned before it is used in the reaction?

1. rub it with sand paper or emery paper to remove the oxide layer
2. rinse it in distilled water and leave it to dry in the air
3. warm it gently in a Bunsen flame until it glows
4. leave it in the open air for several hours before use

Question 3

In the reaction of magnesium with steam, damp mineral wool soaked in water is placed at one end of a heated tube to supply steam. What is the purpose of the mineral wool?

1. to hold and absorb the water so it does not run along the tube
2. to react with the magnesium to make the hydrogen gas
3. to remove impurities from the hydrogen as it passes through
4. to cool the hydrogen gas down before it is collected

Question 4 · Answer A

Step 1: Heating the damp mineral wool boils the water there to produce steam. Step 2: The steam then passes over the magnesium, which must also be heated so it is hot enough to react with the steam. Step 3: So the two places to heat are the damp mineral wool (to make steam) and the magnesium ribbon.

Question 5 · Answer A

Step 1: The steam supplied to the reaction is simply water in the gas state. Step 2: Steam that does not react cools further along the tube and condenses back to liquid water. Step 3: So the colourless liquid that collects is water.

Question 6 · Answer A

Step 1: Before heating, the tube and syringe are full of ordinary air. Step 2: As hydrogen is produced it pushes this air ahead of it, so the first gas collected is a mixture of air and hydrogen. Step 3: So the collected gas is impure because it contains air that was in the apparatus at the start.

Question 7 · Answer A

Step 1: Each temperature change is the highest temperature minus the initial temperature. Step 2: This gives 27.0 − 22.5 = 4.5, 29.0 − 22.5 = 6.5, 30.5 − 22.5 = 8.0, 32.5 − 22.0 = 10.5 and 35.5 − 22.0 = 13.5 °C, each kept to one decimal place. Step 3: So the temperature changes are 4.5, 6.5, 8.0, 10.5 and 13.5 °C.

Question 8 · Answer A

Step 1: A good plot uses a linear scale (equal steps) chosen so the points fill more than half the available grid. Step 2: Every result is plotted accurately, then one smooth line of best fit is drawn through the trend. Step 3: So the correct approach is a linear y-axis scale filling over half the grid, all points plotted accurately and one line of best fit.

Question 9 · Answer A

Step 1: A volume of 45 cm³ is just beyond the plotted results, so the line must be continued (extrapolated) smoothly past the last point. Step 2: From 45 on the volume axis, a vertical line is drawn up to the extended line and then across to the temperature-change axis. Step 3: So the value is found by extending the best-fit line to 45 cm³ and reading across to the temperature-change axis.

Question 10 · Answer A

Step 1: Substitute the values into energy = temperature change × 4.2 × volume of water, giving 13.5 × 4.2 × 15. Step 2: 13.5 × 4.2 = 56.7, and 56.7 × 15 = 850.5. Step 3: So the energy given out is 850.5 J.

Question 11 · Answer A

Step 1: Doubling the mass of lithium chloride from 2.0 g to 4.0 g means twice as much solid dissolves in the same volume of water. Step 2: Twice as much solid releases about twice as much energy to the same mass of water, so the temperature change roughly doubles. Step 3: So the temperature change is estimated at about 9 °C (about double 4.5 °C).

Question 12 · Answer A

Step 1: Dissolving lithium chloride is exothermic, so heat can escape to the surroundings during the experiment. Step 2: Polystyrene is a better insulator than glass, so it reduces this heat loss. Step 3: With less heat lost, the measured temperature change is closer to the true value, so the results are more accurate.

Question 13 · Answer A

Step 1: A measuring cylinder reads volume only to the nearest whole division, which is fairly coarse. Step 2: A burette has a finer scale and is read to a smaller division, so it measures volume more precisely. Step 3: So using a burette is an improvement because it measures the volume of water more accurately.

Question 14 · Answer A

Step 1: The whole point of the investigation is to measure the temperature change caused by the dissolving. Step 2: A water-bath holds the contents at a fixed temperature, so it would stop the temperature from rising. Step 3: So a water-bath is not an improvement because it would keep the temperature constant and prevent the temperature change being observed.

Question 15 · Answer A

Step 1: Iron(II) ions react with sodium hydroxide to form a green precipitate of iron(II) hydroxide. Step 2: Iron(II) hydroxide is insoluble in excess sodium hydroxide, so the precipitate remains when more is added. Step 3: So a green precipitate forms dropwise and stays insoluble in excess.

Question 16 · Answer A

Step 1: Acidified barium nitrate gives a white precipitate of barium sulfate if sulfate ions are present. Step 2: Here no precipitate forms, so there are no sulfate ions in the solution. Step 3: So the observation is no change, showing that sulfate ions are absent.

Question 17 · Answer A

Step 1: Acidified silver nitrate gives a precipitate whose colour identifies the halide: white for chloride, cream for bromide, yellow for iodide. Step 2: This solution contains iodide ions, which give a yellow precipitate. Step 3: So the observation is a yellow precipitate.

Question 18 · Answer A

Step 1: Chlorine is more reactive than iodine, so it displaces iodide ions from solution. Step 2: The displaced iodine colours the solution yellow to brown. Step 3: So adding aqueous chlorine to the iodide solution turns it yellow to brown.

Question 19 · Answer A

Step 1: If ammonium ions were present, warming with sodium hydroxide would give off ammonia, which turns damp red litmus paper blue. Step 2: Here the litmus paper stays red, so no ammonia is produced. Step 3: So the conclusion is that solid K does not contain ammonium ions.

Question 20 · Answer A

Step 1: Warming a nitrate with sodium hydroxide and aluminium foil reduces the nitrate to ammonia. Step 2: Ammonia is the only common gas here that turns damp red litmus paper blue. Step 3: So the gas given off is ammonia.

Question 21 · Answer A

Step 1: A lilac flame shows a potassium ion. Step 2: Giving off ammonia with sodium hydroxide and aluminium foil shows a nitrate ion. Step 3: Combining a potassium ion with a nitrate ion gives potassium nitrate, $\text{KNO}_3$.

Question 22 · Answer A

Step 1: To find a solubility you fix the volume of water (100 cm³) and the temperature (50 °C, maintained), because solubility depends on both. Step 2: Magnesium sulfate is added in portions with stirring until no more dissolves, so the water is saturated at 50 °C. Step 3: The known mass that has dissolved (or filtering off and allowing for the undissolved solid) gives the solubility in g per 100 cm³, so this is the correct plan.