0620/63 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

23 questions · 40 marks · 60 minutes

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1. Question 1(a)

   2 marksNaming fermentation apparatus
   Step 1: The glucose and yeast are held in a conical flask, the standard container for a fermentation that needs a bung and delivery tube. Step 2: A gas syringe collects the carbon dioxide and lets you read its volume directly. Step 3: A measuring cylinder, burette or thermometer would not collect and measure an evolved gas, so A is a conical flask and B is a gas syringe.
   Method:

   Match each labelled item to its job: holding the reaction mixture, and collecting/measuring the gas.
   Examiner tips

   • A gas syringe both collects an evolved gas and reads its volume.

2. Question 1(b)

   1 marksWhy a water-bath is used in fermentation
   Step 1: Fermentation needs a warm temperature, about 25 to 35 degrees C, for the yeast to work. Step 2: A water-bath holds the mixture steadily at this warm temperature. Step 3: Cooling would stop the yeast and boiling would kill it and drive off the ethanol, so the water-bath is used to keep the mixture warm.
   Method:

   Recall the temperature yeast needs and state what the water-bath maintains.
   Examiner tips

   • Fermentation temperature is warm (25-35 degrees C), not hot or cold.

3. Question 1(c)

   1 marksInterpreting a volume-time graph for completion
   Step 1: A reaction is complete when no more product forms, so the volume of gas would level off and stay constant. Step 2: At 90 minutes the volume is still increasing, so gas is still being produced. Step 3: Because the volume has not become constant, fermentation is not yet complete after 90 minutes.
   Method:

   Decide what a completed reaction looks like on a volume-time record, then compare.
   Examiner tips

   • Reaction finished = volume levels off; volume still climbing = not finished.

4. Question 1(d)(i)

   1 marksEffect of carbon dioxide on limewater
   Step 1: Carbon dioxide is the gas produced by fermentation. Step 2: Carbon dioxide reacts with limewater to form insoluble calcium carbonate. Step 3: The fine white solid makes the limewater turn milky (cloudy white), which is the positive test for carbon dioxide.
   Method:

   Recall the limewater test result for carbon dioxide.
   Examiner tips

   • Limewater + carbon dioxide -> milky white.

5. Question 1(d)(ii)

   1 marksObservation showing fermentation is complete
   Step 1: While fermentation continues, carbon dioxide is given off and bubbles pass through the limewater. Step 2: When fermentation is complete, no more carbon dioxide is produced. Step 3: So the observation is that no more bubbles of gas are produced.
   Method:

   Decide what stops happening once no more gas is made.
   Examiner tips

   • Bubbling stops = no more gas = reaction complete.

6. Question 1(e)

   1 marksObtaining ethanol from a fermentation mixture
   Step 1: Ethanol and water are miscible liquids with different boiling points (ethanol boils at about 78 degrees C, water at 100 degrees C). Step 2: Distillation separates liquids by their different boiling points, so the ethanol vaporises first and is condensed and collected. Step 3: Filtration only removes insoluble solids, and crystallisation and chromatography do not separate these two liquids, so distillation is used.
   Method:

   Pick the separation method that works for two miscible liquids with different boiling points.
   Examiner tips

   • Separate two mixed liquids by distillation (different boiling points).

7. Question 2(a)

   3 marksRecording volumes in a rate experiment
   Step 1: The thiosulfate plus water must total 50 cm3 each time (50 + 0 in Experiment 1). Step 2: With 40 cm3 of thiosulfate, the distilled water must be 50 - 40 = 10 cm3. Step 3: The dilute hydrochloric acid is kept the same at 5 cm3 in every experiment, so the recorded values are 10 cm3 of water and 5 cm3 of acid.
   Method:

   Use the constant total volume to find the water, and recall that the acid is unchanged.
   Examiner tips

   • Keep thiosulfate + water constant (50 cm3); keep the acid fixed at 5 cm3.

8. Question 2(b)

   4 marksPlotting results and drawing a best-fit curve
   Step 1: A good graph uses a linear scale that spreads the plotted points over more than half of the grid so the trend is clear. Step 2: Every point must be plotted accurately. Step 3: A single smooth curve of best fit is then drawn through the points, rather than joining them dot-to-dot or using only two of them.
   Method:

   Recall the marks for a good graph: sensible linear scale, accurate points, best-fit line.
   Examiner tips

   • Spread the data over more than half the grid and draw one smooth best-fit line.

9. Question 2(c)(i)

   1 marksCalculating relative rate of reaction
   Step 1: Relative rate = 1 / time. Step 2: With a time of 30 s, relative rate = 1 / 30. Step 3: 1 / 30 = 0.033 (to 2 significant figures), with no units.
   Method:

   Substitute the time into 1/time and evaluate.
   Examiner tips

   • relative rate = 1 / time; 1/30 = 0.033.

10. Question 2(c)(ii)

    1 marksIdentifying the experiment with the greatest rate
    Step 1: Relative rate = 1 / time, so the shortest time gives the greatest rate. Step 2: Experiment 1 has the shortest time (30 s). Step 3: Therefore Experiment 1 has the greatest relative rate of reaction.
    Method:

    Pick the shortest time, since the greatest rate corresponds to the smallest time.
    Examiner tips

    • The fastest reaction is the one that finishes in the least time.

11. Question 2(d)(i)

    2 marksCalculating the volume of water for a fixed total
    Step 1: The thiosulfate and water must add up to 50 cm3. Step 2: Volume of water = 50 - 37. Step 3: 50 - 37 = 13 cm3 of distilled water (the unit cm3 is needed for the second mark).
    Method:

    Subtract the given thiosulfate volume from the constant total volume.
    Examiner tips

    • Subtract the thiosulfate volume from the fixed total of 50 cm3.

12. Question 2(d)(ii)

    2 marksReading a value from a best-fit curve
    Step 1: To read a value off a curve, draw a vertical construction line from 37 cm3 up to the best-fit curve. Step 2: Then draw a horizontal line across to the time axis and read the value. Step 3: Because 37 cm3 lies just below 40 cm3 (37 s), the reading is a little more, about 40 s, and the construction lines must be shown.
    Method:

    Use construction lines from the x value to the curve and across to the y axis.
    Examiner tips

    • Use vertical then horizontal construction lines to read a graph value.

13. Question 2(d)(iii)

    1 marksWhy the total volume is kept constant
    Step 1: Only the thiosulfate concentration is meant to change between experiments. Step 2: Keeping the total volume the same keeps the depth of liquid and the concentration of the acid in the final mixture the same each time. Step 3: This makes the comparison fair, so any change in time is due only to the thiosulfate concentration.
    Method:

    State which variables must stay the same for the test to be fair.
    Examiner tips

    • Control all variables except the one being investigated.

14. Question 2(e)(i)

    1 marksWhy a burette improves accuracy
    Step 1: A burette has finer, more closely spaced graduations than a measuring cylinder. Step 2: This lets the volume be measured to a smaller division and read more precisely. Step 3: So a burette improves the experiment because it measures the volume more accurately.
    Method:

    Compare the precision of the two pieces of apparatus.
    Examiner tips

    • Burette = more accurate volume measurement than a measuring cylinder.

15. Question 2(e)(ii)

    1 marksWhy a volumetric pipette cannot be used
    Step 1: A volumetric pipette is made to deliver one single, fixed volume. Step 2: The thiosulfate volume changes from experiment to experiment (50, 40, 30, 25, 20 cm3). Step 3: One fixed-volume pipette cannot measure all these different volumes, so it cannot be used.
    Method:

    Compare the fixed delivery of a volumetric pipette with the changing volumes needed.
    Examiner tips

    • A volumetric pipette = one fixed volume only.

16. Question 2(f)

    2 marksWhy the beaker size must not change
    Step 1: You judge the time by how long the printed text stays visible through the depth of solution. Step 2: A larger beaker would spread the same volume over a wider base, changing the depth of liquid you look through. Step 3: A different depth changes the time measured for the text to disappear, so the comparison between experiments would no longer be fair.
    Method:

    Link beaker size to depth of liquid, and depth to the time measured.
    Examiner tips

    • Keep the beaker the same so the depth of liquid viewed stays constant.

17. Question 3(a)

    2 marksAluminium ion with sodium hydroxide
    Step 1: Aluminium ions give a white precipitate of aluminium hydroxide when sodium hydroxide is added dropwise. Step 2: Aluminium hydroxide is amphoteric, so it dissolves in excess sodium hydroxide to give a colourless solution. Step 3: So the observations are a white precipitate that dissolves in excess.
    Method:

    Recall the sodium hydroxide test result for aluminium ions, including the excess step.
    Examiner tips

    • Aluminium and zinc hydroxides both dissolve in excess sodium hydroxide.

18. Question 3(b)

    2 marksAluminium ion with aqueous ammonia
    Step 1: Aluminium ions give a white precipitate of aluminium hydroxide with aqueous ammonia added dropwise. Step 2: In excess ammonia the aluminium hydroxide does not dissolve. Step 3: So the precipitate forms and then remains (is insoluble) in excess ammonia.
    Method:

    Recall the ammonia test result for aluminium ions, contrasting it with the sodium hydroxide result.
    Examiner tips

    • Aluminium hydroxide is insoluble in excess ammonia (unlike with sodium hydroxide).

19. Question 3(c)

    1 marksTest for chloride ions
    Step 1: Acidified silver nitrate is the test for halide ions. Step 2: Chloride ions give a white precipitate of silver chloride. Step 3: A cream precipitate would mean bromide and a yellow precipitate iodide, so chloride gives a white precipitate.
    Method:

    Recall the silver nitrate test colour for chloride ions.
    Examiner tips

    • Silver halide colours: chloride white, bromide cream, iodide yellow.

20. Question 3(d)

    1 marksTesting for sulfate ions (negative result)
    Step 1: Acidified barium nitrate gives a white precipitate of barium sulfate only when sulfate ions are present. Step 2: Aluminium chloride contains no sulfate ions. Step 3: With no sulfate present, no precipitate forms, so no change is seen.
    Method:

    Apply the sulfate test and recognise that no sulfate gives no precipitate.
    Examiner tips

    • Barium nitrate gives a white precipitate only if sulfate is present.

21. Question 3(e)

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

    Match the litmus result to the gas that is alkaline.
    Examiner tips

    • Only ammonia turns damp red litmus paper blue.

22. Question 3(f)

    2 marksIdentifying an unknown solid from test results
    Step 1: A light green flame colour indicates barium ions, confirmed by the white precipitate of barium sulfate with sulfuric acid. Step 2: Giving off ammonia when warmed with sodium hydroxide and aluminium foil indicates nitrate ions. Step 3: A solid containing barium ions and nitrate ions is barium nitrate.
    Method:

    Identify the cation from the flame test and the anion from the foil test, then combine them.
    Examiner tips

    • Combine the flame colour (cation) with the foil test (nitrate anion).

23. Question 4

    6 marksPreparing a pure dry sample of an insoluble salt
    Step 1: Because lead(II) chloride is insoluble, it is made by precipitation: dissolve the sodium chloride and lead(II) nitrate separately in distilled water and mix the solutions, so insoluble lead(II) chloride precipitates. Step 2: Filter the mixture to collect the lead(II) chloride as the residue, leaving the soluble sodium nitrate in the filtrate. Step 3: Wash the residue with distilled water to remove the soluble salt, then leave it to dry, giving a pure dry sample.
    Method:

    Use the precipitation route for an insoluble salt: react two soluble solutions, then filter, wash and dry.
    Examiner tips

    • For an insoluble salt: mix solutions, filter, wash the residue, then dry.