IGCSE Chemistry 0620/62 October/November 2025 Past Paper with Answers

Question 1

Small pieces of eggshell, which contain calcium carbonate, are added to dilute hydrochloric acid and the mixture is warmed. What is seen as the calcium carbonate reacts with the acid?

1. effervescence: bubbles of gas are given off
2. the mixture slowly turns deep blue
3. a white precipitate forms and sinks to the bottom
4. no change is seen at all

Question 2

A student must be sure the dilute hydrochloric acid is in excess so that all of the calcium carbonate in the eggshell has reacted. Which check confirms that the acid is in excess?

1. keep adding acid until no more bubbles are produced, or test the mixture with universal indicator and confirm it stays red or orange (still acidic)
2. check that the mixture in the flask has stopped feeling warm to the touch, then assume all of the calcium carbonate has reacted
3. confirm that every piece of the solid present has turned a bright white colour, taking that as proof the acid is in excess
4. weigh the flask before and after the reaction and check that its total mass has increased once the fizzing has finished

Question 3

A student needs to separate the unreacted solid from the liquid by filtration. Which arrangement of apparatus correctly carries out this filtration?

1. a filter funnel holding a filter paper, supported over a beaker, so the liquid passes through and the solid is left on the paper
2. a filter funnel with no filter paper inside it, simply resting in the top of an empty collecting beaker
3. an evaporating basin heated strongly over a roaring Bunsen burner flame until the mixture has all dried up
4. a tall measuring cylinder with a glass stirring rod laid flat across its open top to trap the solid

Question 4 · Answer A

Step 1: The acid is in excess, so leftover dilute hydrochloric acid clings to the residue. Step 2: The reaction makes soluble calcium chloride, which also clings to the residue. Step 3: Both dissolve in the wash water and are rinsed away, so the two substances removed by washing are hydrochloric acid and calcium chloride.

Question 5 · Answer A

Step 1: The calcium carbonate is the part of the 2.00 g sample that reacted, so its mass is 2.00 - 0.11 = 1.89 g. Step 2: The percentage by mass is (mass of calcium carbonate / mass of sample) x 100 = (1.89 / 2.00) x 100. Step 3: This gives 94.5%, so the eggshell is 94.5% calcium carbonate by mass.

Question 6 · Answer A

Step 1: A titre is the final (end-point) burette reading minus the initial reading, recorded to one decimal place. Step 2: Experiment 1: first end-point 21.2 - 1.6 = 19.6 cm³, second end-point 41.0 - 1.6 = 39.4 cm³. Step 3: Experiment 2: first end-point 10.2 - 0.4 = 9.8 cm³, second end-point 20.1 - 0.4 = 19.7 cm³, giving titres of 19.6, 39.4, 9.8 and 19.7 cm³.

Question 7 · Answer A

Step 1: The flask was used in the previous experiment, so traces of the old solution could be left inside. Step 2: Rinsing with distilled water washes these contaminants away so they cannot affect the next result. Step 3: The rinse does not react with anything or change concentrations, so its purpose is simply to clean the flask and remove contaminants.

Question 8 · Answer A

Step 1: Distilled water left clinging inside the burette mixes with solution C and dilutes it, so the solution C run out is less concentrated. Step 2: A less concentrated alkali contains fewer moles per cm³, so more of it is needed to neutralise the same acid. Step 3: Therefore the volume added to reach the first end-point would be greater, because the leftover water dilutes solution C.

Question 9 · Answer A

Step 1: The end-point is judged from a colour change of the indicator in the flask. Step 2: A plain white background makes that colour stand out, so the change is easier and clearer to see. Step 3: The tile does not warm the flask or affect the burette reading, so its purpose is to make the colour change easier to see.

Question 10 · Answer A

Step 1: The same amount of acid in solution B is neutralised in both experiments, so the amount of sodium hydroxide reacting is the same each time. Step 2: That same amount is supplied by 39.4 cm³ of C but only 19.7 cm³ of D, so D is more concentrated because it delivers the same amount in a smaller volume. Step 3: Since 39.4 is about twice 19.7, solution D is about twice the concentration of solution C.

Question 11 · Answer A

Step 1: Doubling the volume of solution B from 25 cm³ to 50 cm³ doubles the amount of acid to be neutralised. Step 2: Twice the acid needs twice the volume of solution C to reach the same first end-point. Step 3: So the volume of C is 19.6 x 2 = 39.2 cm³ (the unit is cm³).

Question 12 · Answer A

Step 1: Doubling solution B to 50 cm³ doubles the volume of C needed for the second end-point, from 39.4 cm³ to about 78.8 cm³. Step 2: The burette only holds 50 cm³, so it cannot deliver 78.8 cm³ in one filling. Step 3: The burette would have to be refilled part-way, which is awkward and less accurate, so the problem is that the volume needed is more than the burette can hold.

Question 13 · Answer A

Step 1: The titre depends on the amount (moles) of acid in solution B, not on its temperature. Step 2: Warming the solution does not add or remove acid, so the number of moles of acid stays the same. Step 3: The same amount of acid needs the same amount of solution D to neutralise it, so there is no change in the volume of D required.

Question 14 · Answer A

Step 1: A measuring cylinder measures volume only roughly, so using it for the fixed 25 cm³ of solution B limits the accuracy. Step 2: A volumetric pipette delivers a fixed 25 cm³ much more accurately. Step 3: Swapping the measuring cylinder for a volumetric pipette therefore improves the accuracy of the results.

Question 15 · Answer A

Step 1: Warming an ammonium salt with sodium hydroxide releases ammonia gas. Step 2: Ammonia is an alkaline gas, and an alkaline gas turns damp red litmus paper blue. Step 3: Ammonia does not bleach litmus or pop with a splint, so the observation is that the damp red litmus paper turns blue.

Question 16 · Answer A

Step 1: Warming an ammonium salt with sodium hydroxide produces ammonia gas. Step 2: Ammonia is the only common gas that is alkaline, which is why it turns damp red litmus paper blue. Step 3: Hydrogen, carbon dioxide and chlorine do not turn damp red litmus blue, so the gas is ammonia.

Question 17 · Answer A

Step 1: Acidified barium nitrate is the test for sulfate ions. Step 2: Sulfate ions react with the barium ions to give a white precipitate of insoluble barium sulfate. Step 3: A cream or brown precipitate would mean other ions, so sulfate ions give a white precipitate.

Question 18 · Answer A

Step 1: Acidified potassium manganate(VII) is decolourised (purple to colourless) only when a reducing agent is present. Step 2: Sulfate ions are not a reducing agent (unlike sulfite ions), so there is nothing to decolourise the manganate(VII). Step 3: The purple colour therefore remains, appearing pink/purple in the diluted few drops, so the solution is not decolourised.

Question 19 · Answer A

Step 1: A yellow flame is the flame colour of sodium ions, so one ion is sodium, Na⁺. Step 2: A white precipitate that dissolves in excess of BOTH sodium hydroxide and ammonia (each to a colourless solution) is given by zinc, Zn²⁺; aluminium would not dissolve in excess ammonia. Step 3: A yellow precipitate with acidified silver nitrate is iodide, I⁻ (chloride is white, bromide cream), so the three ions are sodium, zinc and iodide.

Question 20 · Answer A

Step 1: For a fair test only the alcohol should change, so use the same fixed volume of water (measured with a measuring cylinder, burette or pipette) and record its starting temperature each time. Step 2: Heat the water with one alcohol burner and measure the heating in a comparable way: either time how long it takes to reach a set temperature, or heat for a fixed time and find the temperature rise. Step 3: Repeat under identical conditions for all three alcohols; the alcohol with the shortest time (or the largest temperature rise) heats the water fastest.