0620/53 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

20 questions · 40 marks · 75 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1(a)

   4 marksRecording reaction times as concentration changes
   Step 1: Adding distilled water lowers the concentration of the sodium thiosulfate while the total volume stays the same. Step 2: A lower concentration means fewer thiosulfate particles in the same volume, so collisions are less frequent and the reaction is slower, making the cloudiness take longer to hide the text. Step 3: So the time increases as the volume of sodium thiosulfate decreases, and each time is recorded in whole seconds.
   Method:

   Link the diluted concentration to collision frequency, then to rate, then to the time recorded.
   Examiner tips

   • Concentration down = rate down = time up.

2. Question 1(b)

   4 marksPlotting a graph and drawing a curve of best fit
   Step 1: A good graph uses a linear scale (equal steps) chosen so the plotted points fill more than half of the available grid. Step 2: Every result is plotted accurately, then a single smooth curve 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 curve of best fit.
   Method:

   Recall the rules for a good graph: linear scale, points spread out, single curve of best fit.
   Examiner tips

   • Choose a linear scale so the points use over half the grid; draw one curve of best fit.

3. Question 1(c)(i)

   1 marksCalculating relative rate of reaction
   Step 1: The relative rate is 1 divided by the time taken. Step 2: Substituting the Experiment 1 time gives 1 ÷ 30 = 0.033. Step 3: The question says to give no units, so the relative rate of reaction is 0.033.
   Method:

   Substitute the Experiment 1 time into relative rate = 1 ÷ time.
   Examiner tips

   • Relative rate = 1 ÷ time taken; no units required.

4. Question 1(c)(ii)

   1 marksIdentifying the experiment with the greatest rate
   Step 1: The relative rate is 1 ÷ time, so the greatest rate goes with the shortest time. Step 2: Experiment 1 uses the most sodium thiosulfate (50 cm³), so it has the highest concentration and reacts fastest, giving the shortest time. Step 3: So the relative rate of reaction is greatest in Experiment 1.
   Method:

   Match the greatest rate to the experiment with the highest concentration and shortest time.
   Examiner tips

   • Highest concentration = shortest time = greatest relative rate.

5. Question 1(d)(i)

   2 marksCalculating a volume to keep the total constant
   Step 1: The sodium thiosulfate and the distilled water must add up to a fixed total of 50 cm³. Step 2: So the water needed is 50 cm³ − 37 cm³ = 13 cm³. Step 3: Keeping the unit, the volume of distilled water is 13 cm³.
   Method:

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

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

6. Question 1(d)(ii)

   2 marksReading a value by interpolation from a curve of best fit
   Step 1: To read an intermediate value, find 37 cm³ on the volume axis. Step 2: Draw a vertical construction line up to the curve of best fit, then a horizontal line across to the time axis. Step 3: The value where the horizontal line meets the time axis is the answer, so you draw a vertical line up from 37 cm³ then read horizontally across to the time axis.
   Method:

   Use construction lines to interpolate the time at the stated volume.
   Examiner tips

   • Interpolation: up to the curve, then across to the axis.

7. Question 1(d)(iii)

   1 marksWhy the total volume must be kept constant
   Step 1: Only the concentration of the sodium thiosulfate should change between experiments, so everything else must be controlled. Step 2: Keeping the total volume fixed keeps the depth of liquid you look down through the same, and keeps the acid diluted to the same concentration in the final mixture. Step 3: So the total volume is kept constant so the depth and the acid concentration are the same every time, making it a fair test.
   Method:

   Identify the variables a constant total volume controls (depth and acid concentration).
   Examiner tips

   • Constant total volume = constant depth and constant diluted acid concentration.

8. Question 1(e)(i)

   1 marksChoosing more accurate apparatus to measure a volume
   Step 1: The accuracy of a result depends on how precisely the volumes are measured. Step 2: A burette has finer graduations than a measuring cylinder, so it measures a volume of liquid more accurately. Step 3: So using a burette is an improvement because it is more accurate than a measuring cylinder.
   Method:

   Compare the precision of a burette with a measuring cylinder.
   Examiner tips

   • Burette = finer graduations = more accurate measurement.

9. Question 1(e)(ii)

   1 marksWhy a volumetric pipette is unsuitable for a varying volume
   Step 1: A volumetric pipette is designed to deliver a single fixed volume very accurately. Step 2: In this investigation the volume of sodium thiosulfate is changed in every experiment (50, 40, 30, 25, 20 cm³), so a fixed-volume pipette cannot measure them all. Step 3: So a volumetric pipette is unsuitable because the thiosulfate volume is not constant.
   Method:

   Match the fixed-volume nature of a pipette against the need for a varying volume.
   Examiner tips

   • Fixed-volume apparatus cannot measure a volume that changes between experiments.

10. Question 1(f)

    2 marksWhy the size of the beaker must not change
    Step 1: You judge the end point by looking down through the liquid until the text disappears, so the depth of liquid affects how long that takes. Step 2: Putting the same volume into a larger beaker spreads it out and changes the depth of liquid you look through. Step 3: A different depth changes the time measured, so the size of the beaker must be kept the same.
    Method:

    Link beaker size to the depth of liquid looked through and then to the measured time.
    Examiner tips

    • Same beaker = same depth = a fair comparison of times.

11. Question 2(a)

    2 marksReaction of an aluminium salt with sodium carbonate
    Step 1: Aluminium ions are acidic in solution, so adding sodium carbonate produces a white precipitate of aluminium hydroxide. Step 2: At the same time carbon dioxide is released, so effervescence (fizzing) is seen. Step 3: So adding sodium carbonate to solution L gives a white precipitate together with effervescence.
    Method:

    Apply the reaction of an acidic aluminium salt with sodium carbonate.
    Examiner tips

    • Aluminium ion + carbonate → white precipitate plus carbon dioxide.

12. Question 2(b)

    2 marksEffect of sodium hydroxide on an aluminium salt
    Step 1: Adding sodium hydroxide to an aluminium salt first gives a white precipitate of aluminium hydroxide. Step 2: Aluminium hydroxide is amphoteric, so it dissolves in excess sodium hydroxide to give a colourless solution. Step 3: So a white precipitate forms dropwise and dissolves in excess.
    Method:

    Match the white-precipitate-soluble-in-excess result to an aluminium ion.
    Examiner tips

    • White precipitate, soluble in excess sodium hydroxide = aluminium or zinc.

13. Question 2(c)

    2 marksEffect of ammonia on an aluminium salt
    Step 1: Adding ammonia to an aluminium salt gives a white precipitate of aluminium hydroxide. Step 2: Aluminium hydroxide is insoluble in excess ammonia, so the precipitate remains. Step 3: So a white precipitate forms dropwise and remains insoluble in excess ammonia.
    Method:

    Distinguish aluminium from zinc by the insoluble-in-excess-ammonia result.
    Examiner tips

    • Aluminium: white precipitate insoluble in excess ammonia (but soluble in excess sodium hydroxide).

14. Question 2(d)

    1 marksSilver nitrate test giving a white precipitate
    Step 1: Acidified silver nitrate gives a precipitate whose colour identifies the halide: white for chloride, cream for bromide, yellow for iodide. Step 2: Solution L gives a white precipitate, which corresponds to a chloride ion. Step 3: So the observation is a white precipitate.
    Method:

    Match the silver nitrate precipitate colour to the halide ion present.
    Examiner tips

    • Silver nitrate: chloride white, bromide cream, iodide yellow.

15. Question 2(e)

    2 marksIdentifying a solid from its cation and anion tests
    Step 1: A white precipitate with sodium hydroxide that is soluble in excess but insoluble in excess ammonia shows an aluminium ion (this distinguishes it from zinc, which dissolves in excess ammonia too). Step 2: A white precipitate with acidified silver nitrate shows a chloride ion. Step 3: Combining an aluminium ion with chloride ions gives aluminium chloride, $\text{AlCl}_3$.
    Method:

    Deduce the cation and anion from the tests, then name the compound.
    Examiner tips

    • Identify the cation and the anion separately, then combine them.

16. Question 2(f)

    1 marksFlame test colour for barium
    Step 1: Each metal ion gives a characteristic flame colour. Step 2: Barium ions give a light green flame. Step 3: So the flame colour for solid M is light green.
    Method:

    Recall the flame colour produced by barium ions.
    Examiner tips

    • Flame colours: Na yellow, K lilac, Ca orange-red, Ba light green.

17. Question 2(g)(i)

    2 marksTesting for nitrate ions with sodium hydroxide and aluminium
    Step 1: Warming a nitrate with sodium hydroxide and aluminium foil reduces the nitrate to ammonia. Step 2: Bubbles of ammonia gas form (effervescence) and the gas turns damp red litmus paper blue. Step 3: So effervescence is seen and the gas turns damp red litmus paper blue, confirming ammonia and so a nitrate ion.
    Method:

    Apply the nitrate test and the test for ammonia gas.
    Examiner tips

    • Nitrate test: warm with NaOH + Al foil → ammonia → damp red litmus turns blue.

18. Question 2(g)(ii)

    1 marksReaction of a barium salt with sulfuric acid
    Step 1: Barium ions react with sulfate ions from the sulfuric acid to form barium sulfate, which is insoluble. Step 2: The insoluble barium sulfate appears as a white precipitate. Step 3: So adding dilute sulfuric acid to solution M gives a white precipitate.
    Method:

    Apply the rule that barium sulfate is an insoluble white solid.
    Examiner tips

    • Barium sulfate is insoluble, so barium + sulfuric acid gives a white precipitate.

19. Question 2(h)

    2 marksIdentifying a solid from flame and ion tests
    Step 1: A light green flame shows a barium ion, and the white precipitate with sulfuric acid confirms barium. Step 2: Producing ammonia when warmed with sodium hydroxide and aluminium foil shows a nitrate ion. Step 3: Combining a barium ion with nitrate ions gives barium nitrate, $\text{Ba(NO}_3)_2$.
    Method:

    Deduce the cation from the flame test and the anion from the nitrate test, then name the salt.
    Examiner tips

    • Identify the metal from the flame test and the anion from the nitrate test.

20. Question 3

    6 marksPreparing an insoluble salt by precipitation
    Step 1: Dissolve each solid separately in distilled water, then mix the solutions so the insoluble lead(II) chloride forms as a precipitate while the soluble sodium nitrate stays in solution. Step 2: Filter the mixture to collect the lead(II) chloride as the residue, then wash the residue with distilled water to remove the soluble sodium nitrate. Step 3: Finally dry the solid, giving a pure, dry sample of lead(II) chloride.
    Method:

    Design a precipitation preparation: dissolve, mix, filter the residue, wash and dry.
    Examiner tips

    • Insoluble salt = precipitate, filter as the residue, wash, then dry.