0620/52 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

19 questions · 40 marks · 75 minutes

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1. Question 1(a)

   5 marksTrend in temperature change as the volume of water changes
   Step 1: Each experiment releases the same amount of energy because the same 2.0 g of lithium chloride dissolves every time. Step 2: That fixed energy warms the water; a smaller volume of water has less mass to share the energy, so its temperature rises more. Step 3: So the temperature change increases as the volume of water decreases.
   Method:

   Link the fixed energy released to the mass of water being heated, then to the temperature change.
   Examiner tips

   • Same energy ÷ less water = bigger temperature change.

2. Question 1(b)

   4 marksPlotting temperature change against volume of water
   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 line of best fit is drawn through the trend. Step 3: So the correct approach is a linear y-axis scale with the points over half the grid, all points plotted accurately and one line of best fit.
   Method:

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

   • Choose a scale so the points use more than half the grid; draw one best-fit line.

3. Question 1(c)

   3 marksExtrapolating a best-fit line beyond the plotted points
   Step 1: The value at 45 cm³ lies beyond the plotted points, so the line must be extended (extrapolated) smoothly to 45 cm³. Step 2: Draw a construction line up from 45 cm³ to the extended line, then across to the temperature-change axis. Step 3: Read the value where it meets the axis and give it in °C, so you extend the line to 45 cm³ then read across.
   Method:

   Continue the line past the plotted points and read off the value at 45 cm³ with its unit.
   Examiner tips

   • Extrapolation: extend the line beyond the data, then read off with the correct unit.

4. Question 1(d)

   1 marksCalculating the energy given out on dissolving
   Step 1: Substitute the values into energy = temperature change × 4.2 × volume of water. Step 2: That is 20.0 × 4.2 × 15. Step 3: 20.0 × 4.2 = 84, and 84 × 15 = 1260, so the energy given out is 1260 J.
   Method:

   Substitute the temperature change, 4.2 and the 15 cm³ volume into the given equation.
   Examiner tips

   • Use all three factors: temperature change × 4.2 × volume of water.

5. Question 1(e)

   2 marksEffect of doubling the mass of solid on the temperature change
   Step 1: The energy released is proportional to the mass of lithium chloride that dissolves. Step 2: Doubling the mass from 2.0 g to 4.0 g doubles the energy given out, and the same volume of water now receives twice the energy. Step 3: So the temperature change roughly doubles, from 10.0 °C to about 20.0 °C.
   Method:

   Use proportionality: double the mass, double the energy, so double the temperature change.
   Examiner tips

   • Energy released is proportional to the mass dissolved.

6. Question 1(f)

   2 marksWhy a polystyrene cup improves accuracy
   Step 1: Dissolving lithium chloride gives out heat, and some of that heat can escape to the surroundings. Step 2: Polystyrene is a better insulator than glass, so it reduces the heat lost from the cup. Step 3: With less heat lost, the measured temperature change is closer to the true value, so the results are more accurate.
   Method:

   Link the insulating property of polystyrene to reduced heat loss and a more accurate temperature change.
   Examiner tips

   • Insulation reduces heat loss and improves the accuracy of the temperature change.

7. Question 1(g)(i)

   1 marksWhy a burette is more accurate than a measuring cylinder
   Step 1: A burette has finer graduations than a measuring cylinder, so the volume can be read to a smaller division. Step 2: This means the volume of water is measured more precisely. Step 3: So using a burette is an improvement because it measures the volume of water more accurately.
   Method:

   Recall that a burette reads volume to a smaller division, giving greater accuracy.
   Examiner tips

   • Burette = finer graduations = more accurate volume.

8. Question 1(g)(ii)

   1 marksWhy a water-bath is not an improvement for this experiment
   Step 1: The whole point of the experiment is to measure the temperature change as the lithium chloride dissolves. Step 2: A water-bath holds the contents at a fixed temperature, so it would cancel out the very temperature change being measured. Step 3: So a water-bath is not an improvement because it would keep the temperature constant.
   Method:

   Recognise that fixing the temperature defeats an experiment whose aim is to measure a temperature change.
   Examiner tips

   • Do not hold the temperature constant when the temperature change is the measurement.

9. Question 2(a)

   2 marksEffect of sodium hydroxide on an iron(II) salt solution
   Step 1: Adding sodium hydroxide to an iron(II) salt gives a green precipitate of iron(II) hydroxide. Step 2: Iron(II) hydroxide is insoluble in excess sodium hydroxide, so the precipitate stays when more is added. Step 3: So a green precipitate forms dropwise and it remains insoluble in excess.
   Method:

   Match the green-precipitate-insoluble-in-excess result to an iron(II) ion.
   Examiner tips

   • Green ppt, insoluble in excess NaOH = iron(II).

10. Question 2(b)

    2 marksChlorine displacing iodine, confirmed by starch
    Step 1: Chlorine is more reactive than iodine, so it displaces iodide ions to form iodine, which colours the solution brown. Step 2: Iodine gives a blue-black colour with starch. Step 3: So the solution turns brown with chlorine and then blue-black when starch is added.
    Method:

    Apply halogen reactivity (chlorine displaces iodine) and the starch test for iodine.
    Examiner tips

    • More reactive halogen displaces a less reactive one; starch detects iodine as blue-black.

11. Question 2(c)

    1 marksSilver nitrate test giving a yellow 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 J contains iodide ions, which give a yellow precipitate. Step 3: So the observation is a yellow precipitate.
    Method:

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

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

12. Question 2(d)

    1 marksReaction of an iron(II) salt with sodium carbonate
    Step 1: Iron(II) ions react with carbonate ions to form iron(II) carbonate, which is insoluble. Step 2: Iron(II) compounds are green, so the precipitate is green. Step 3: So adding sodium carbonate to solution J gives a green precipitate.
    Method:

    Apply that iron(II) carbonate is an insoluble green solid.
    Examiner tips

    • Iron(II) salts give green insoluble precipitates with carbonate.

13. Question 2(e)

    2 marksIdentifying a solid from its test results
    Step 1: A green precipitate with sodium hydroxide, insoluble in excess, shows an iron(II) ion. Step 2: A yellow precipitate with silver nitrate shows an iodide ion. Step 3: Combining an iron(II) ion with iodide ions gives iron(II) iodide, $\text{FeI}_2$.
    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.

14. Question 2(f)

    1 marksFlame test colour for potassium
    Step 1: Each metal ion gives a characteristic flame colour. Step 2: Potassium ions give a lilac flame. Step 3: So the flame colour for solid K is lilac.
    Method:

    Recall the flame colour produced by potassium ions.
    Examiner tips

    • Flame colours: Na yellow, K lilac, Ca orange-red, Cu blue-green.

15. Question 2(g)(i)

    1 marksTesting for ammonium ions with warm sodium hydroxide
    Step 1: Warming a solution with sodium hydroxide releases ammonia only if ammonium ions are present, and ammonia turns damp red litmus blue. Step 2: Here the litmus stays red, so no ammonia is given off. Step 3: So the observation is no change, with the red litmus paper staying red.
    Method:

    Apply the warm sodium hydroxide test for ammonium and interpret an unchanged litmus.
    Examiner tips

    • Damp red litmus turns blue only if ammonia is released.

16. Question 2(g)(ii)

    1 marksConclusion from a negative ammonium test
    Step 1: Ammonium ions release ammonia when warmed with sodium hydroxide, turning damp red litmus blue. Step 2: No ammonia was given off, so the litmus stayed red. Step 3: So the conclusion is that solid K does not contain ammonium ions.
    Method:

    Interpret the negative ammonia result as the absence of ammonium ions.
    Examiner tips

    • A negative ammonia test rules out ammonium ions.

17. Question 2(g)(iii)

    2 marksTest for nitrate ions using aluminium and sodium hydroxide
    Step 1: Aluminium with warm sodium hydroxide reduces nitrate ions to ammonia gas, so bubbles of gas are seen. Step 2: Ammonia is alkaline and turns damp red litmus paper blue. Step 3: So the observations are effervescence and the gas turning damp red litmus paper blue, showing nitrate ions are present.
    Method:

    Apply the aluminium and sodium hydroxide test for nitrate, which gives ammonia.
    Examiner tips

    • Nitrate test: warm with aluminium + sodium hydroxide gives ammonia (red litmus blue).

18. Question 2(h)

    2 marksIdentifying a solid from flame and nitrate tests
    Step 1: A lilac flame shows a potassium ion. Step 2: Producing ammonia with aluminium and warm sodium hydroxide shows a nitrate ion. Step 3: Combining a potassium ion with a nitrate ion gives potassium nitrate, $\text{KNO}_3$.
    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.

19. Question 3

    6 marksPlanning a measurement of solubility at a fixed temperature
    Step 1: Use a measured 100 cm³ of water in a suitable container and heat it to 50 °C, holding it at that temperature so the result is for the stated temperature. Step 2: Add magnesium sulfate in portions with stirring until no more dissolves, so the water is saturated. Step 3: Find the mass that dissolved (for example by filtering or weighing the excess), then state it as grams per 100 cm³ of water, which is the solubility.
    Method:

    Design a controlled method that saturates a known volume of water at the stated temperature and finds the mass dissolved.
    Examiner tips

    • Control the volume of water and the temperature, saturate, then find mass dissolved per 100 cm³.