0620/61 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

22 questions · 40 marks · 60 minutes

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1. Question 1(a)

   1 marksNaming the reaction vessel in a gas preparation
   Step 1: A vessel with a wide flat base, sloping sides and a narrow neck is a conical flask, which is used to hold a reaction mixture. Step 2: Its narrow neck makes it easy to fit a delivery tube and to swirl the contents. Step 3: A measuring cylinder and pipette only measure volumes and a crucible is for strong heating of solids, so the vessel is a conical flask.
   Method:

   Match the described shape to the standard piece of laboratory glassware.
   Examiner tips

   • Conical flask = flat base, sloping sides, narrow neck.

2. Question 1(b)

   1 marksIdentifying the reagent in a gas preparation
   Step 1: Chlorine is prepared in the laboratory by warming manganese(IV) oxide with hot concentrated hydrochloric acid. Step 2: The manganese(IV) oxide oxidises the chloride ions in the acid to chlorine gas. Step 3: Sulfuric acid, sodium hydroxide and water do not provide the chloride ions needed, so the corrosive liquid Y is concentrated hydrochloric acid.
   Method:

   Identify which liquid supplies the chloride ions that are oxidised to chlorine.
   Examiner tips

   • Manganese(IV) oxide + concentrated hydrochloric acid (warm) -> chlorine.

3. Question 1(c)

   1 marksChoosing apparatus to heat a reaction
   Step 1: Heating a flask in the laboratory is done with a Bunsen burner, which provides a hot, adjustable flame. Step 2: The flame supplies the energy needed to drive off the chlorine gas. Step 3: A thermometer only measures temperature, a measuring cylinder measures volume and a filter funnel separates solids, so the heat source is a Bunsen burner.
   Method:

   Pick the apparatus whose job is to provide heat.
   Examiner tips

   • A Bunsen burner is the usual way to heat a flask in the lab.

4. Question 1(d)

   1 marksCorrecting a gas-collection error for a dense gas
   Step 1: Chlorine is more dense than air, so it sinks and must be collected in a tube held the right way up (mouth upwards), pushing the air out of the top. Step 2: This is downward delivery, also called collecting by upward displacement of air. Step 3: An inverted tube or collection over water (chlorine is soluble) would not work, so the tube is turned the right way up for downward delivery.
   Method:

   Use the density of the gas to choose the correct collection method.
   Examiner tips

   • Denser-than-air gas -> downward delivery into an upright tube.

5. Question 1(e)

   1 marksReason for using a fume cupboard
   Step 1: Chlorine is a toxic (poisonous) gas that is harmful if breathed in. Step 2: A fume cupboard draws the gas away and stops it building up where it could be inhaled. Step 3: Chlorine is not flammable or cold and is actually reactive, so the reason for the fume cupboard is that chlorine is toxic.
   Method:

   Link the hazard of the gas to the purpose of a fume cupboard.
   Examiner tips

   • Use a fume cupboard whenever a toxic gas such as chlorine is made.

6. Question 2(a)

   5 marksRecording titration readings and finding the titre
   Step 1: Burette readings are recorded to one decimal place and the titre is the final reading minus the initial reading; the fixed acid volume is 25.0 cm³ each time. Step 2: The sodium hydrogencarbonate neutralises some of the hydrochloric acid before the titration, so more of it leaves less acid for the sodium hydroxide to neutralise. Step 3: Therefore the titre falls steadily as the mass of sodium hydrogencarbonate increases (45.5, 40.0, 34.6, 23.8, 13.0 cm³).
   Method:

   Apply the burette-reading rules, then reason how added sodium hydrogencarbonate changes the acid left to titrate.
   Examiner tips

   • Read a burette to 1 dp; titre = final reading - initial reading.

7. Question 2(b)

   4 marksPlotting titration results and drawing a best-fit line
   Step 1: A linear y-axis scale chosen so the points fill more than half the grid makes the plot easy to read and keeps plotting error small. Step 2: Each of the five points is then plotted as accurately as possible. Step 3: A single straight line of best fit is drawn so the points are evenly balanced about it, rather than joined dot-to-dot.
   Method:

   Apply the rules for choosing a scale, plotting points and drawing a line of best fit.
   Examiner tips

   • Use a linear scale, plot every point, and draw one straight best-fit line.

8. Question 2(c)

   3 marksExtrapolating a titration line to find a mass
   Step 1: When all the hydrochloric acid has been neutralised by the sodium hydrogencarbonate, no sodium hydroxide is needed, so the titre is 0. Step 2: On the graph the titre is 0 where the line meets the x-axis, so extrapolate the line to the x-axis until the titre reaches 0. Step 3: The mass of sodium hydrogencarbonate read off at that point (about 2.1 g) is the answer, and it is a mass so the unit is grams.
   Method:

   Recognise that zero titre marks complete neutralisation, then read the matching mass from the extrapolated line.
   Examiner tips

   • Titre = 0 means no sodium hydroxide needed, i.e. all the acid is already neutralised.

9. Question 2(d)(i)

   1 marksChoosing a pipette over a measuring cylinder
   Step 1: The same fixed volume, 25.0 cm³, of acid is needed in every experiment so that the results can be compared fairly. Step 2: A volumetric pipette delivers this fixed volume more accurately (to a finer precision) than a measuring cylinder. Step 3: The glassware does not react with the acid and both can hold acid, so the reason is the greater accuracy of the pipette.
   Method:

   Compare the accuracy of the two pieces of apparatus for delivering a fixed volume.
   Examiner tips

   • Pipette = accurate fixed volume; measuring cylinder = rough volume.

10. Question 2(d)(ii)

    1 marksWhy a pipette cannot replace a burette
    Step 1: The volume of sodium hydroxide needed to reach the end-point is different in each experiment (the titre changes as more sodium hydrogencarbonate is used). Step 2: A volumetric pipette is made to deliver only one fixed volume, so it cannot measure these varying volumes. Step 3: A burette can deliver and measure any volume up to its capacity, so the burette is used and a pipette cannot replace it.
    Method:

    Match the changing volume requirement to the apparatus that can measure variable volumes.
    Examiner tips

    • Varying volumes need a burette; a pipette gives only one fixed volume.

11. Question 2(e)

    2 marksWhy universal indicator is unsuitable for a titration
    Step 1: A titration needs an indicator that changes colour sharply at the end-point so the exact volume can be read. Step 2: Universal indicator passes through many colours gradually as the pH changes, so there is no single sharp change. Step 3: Because the colour changes gradually, the precise end-point cannot be judged, so universal indicator is unsuitable and a single-colour-change indicator is used instead.
    Method:

    Compare the gradual colour range of universal indicator with the sharp change a titration needs.
    Examiner tips

    • A titration needs a sharp single colour change, which universal indicator cannot give.

12. Question 2(f)

    2 marksEffect of doubling the alkali concentration on the line
    Step 1: The amount of acid to be neutralised in each experiment is unchanged. Step 2: If the sodium hydroxide is twice as concentrated, each cm³ neutralises twice as much acid, so only about half the volume is needed to reach the end-point. Step 3: Every titre is therefore roughly halved, giving a straight line lying below the original line at every mass (it ends on the x-axis at about the same mass, 2.0 to 2.2 g, where all the acid is neutralised).
    Method:

    Reason how concentration and volume trade off for a fixed amount of acid.
    Examiner tips

    • Double the concentration -> about half the volume -> line below the original.

13. Question 3(a)(i)

    1 marksFlame colour of lithium
    Lithium ions colour a Bunsen flame red, which is the flame-test result that tells lithium apart from sodium (yellow), potassium (lilac) and copper(II) (blue-green).
    Method:

    Recall the flame colour associated with lithium ions.
    Examiner tips

    • Lithium -> red flame; sodium -> yellow; potassium -> lilac.

14. Question 3(a)(ii)

    1 marksWhy a yellow Bunsen flame is unsuitable for a flame test
    Step 1: A flame test needs a hot flame so that the metal ions are heated enough to give their characteristic colour. Step 2: A yellow (luminous) Bunsen flame is cooler than the roaring blue flame, so it is not hot enough to produce a clear colour. Step 3: Its own yellow colour would also mask the result, so a hot blue flame is used instead because the yellow flame is not hot enough.
    Method:

    Link the temperature of the yellow flame to its failure to produce a flame colour.
    Examiner tips

    • Use the roaring blue flame; the yellow flame is cooler (not hot enough).

15. Question 3(b)(i)

    1 marksSilver nitrate test for chloride ions
    Step 1: Halide ions are tested by acidifying with dilute nitric acid and then adding aqueous silver nitrate. Step 2: Chloride ions give a white precipitate of silver chloride. Step 3: A cream precipitate would mean bromide and a yellow one iodide, so chloride ions give a white precipitate.
    Method:

    Recall the silver nitrate halide test and match the colour to chloride.
    Examiner tips

    • Chloride -> white, bromide -> cream, iodide -> yellow with silver nitrate.

16. Question 3(b)(ii)

    1 marksSodium hydroxide added to a lithium salt
    Step 1: Sodium hydroxide forms insoluble metal hydroxide precipitates with many metal ions, but the hydroxides of Group I metals are soluble. Step 2: Lithium is a Group I metal, so lithium hydroxide stays dissolved and no precipitate forms. Step 3: Therefore no change is seen when sodium hydroxide is added to a lithium solution.
    Method:

    Use the solubility of Group I hydroxides to predict no precipitate.
    Examiner tips

    • Lithium (Group I) gives no precipitate with sodium hydroxide.

17. Question 3(c)

    1 marksConclusion from a test for water
    Step 1: Anhydrous cobalt(II) chloride paper turning from blue to pink is the test for water, and condensation at the top of the tube is liquid water. Step 2: Water is driven off when solid K is heated, which means the water was held in the solid. Step 3: A solid that contains water of crystallisation is described as hydrated, so the conclusion is that solid K is hydrated.
    Method:

    Recognise the water test, then conclude the solid contained water of crystallisation.
    Examiner tips

    • Cobalt(II) chloride paper blue -> pink = water present; solid is hydrated.

18. Question 3(d)(i)

    2 marksTwo cations giving a green precipitate with ammonia
    Step 1: A green precipitate with aqueous ammonia points to a green metal hydroxide. Step 2: Both iron(II) and chromium(III) give green precipitates with aqueous ammonia that are insoluble in excess. Step 3: Iron(III) gives a brown precipitate, copper(II) a blue one, and aluminium and zinc give white precipitates, so the two cations are iron(II) and chromium(III).
    Method:

    Match the green, excess-insoluble precipitate to the two cations that give it.
    Examiner tips

    • Green hydroxide insoluble in excess ammonia = iron(II) or chromium(III).

19. Question 3(d)(ii)

    2 marksConfirming chromium(III) versus iron(II)
    Step 1: Chromium(III) hydroxide is amphoteric, so its green precipitate dissolves in excess sodium hydroxide, while iron(II) hydroxide is not and stays insoluble. Step 2: So adding excess aqueous sodium hydroxide separates them: a precipitate that dissolves is chromium(III). Step 3: A precipitate that does not dissolve is iron(II), so the dissolving behaviour in excess sodium hydroxide confirms which cation is present.
    Method:

    Use the different solubility of the two green hydroxides in excess sodium hydroxide.
    Examiner tips

    • Excess sodium hydroxide dissolves chromium(III) hydroxide but not iron(II) hydroxide.

20. Question 3(e)

    1 marksIdentifying a gas that turns limewater milky
    Step 1: The test that turns limewater milky is the test for carbon dioxide. Step 2: Carbon dioxide reacts with limewater to form insoluble calcium carbonate, which makes it cloudy (milky). Step 3: Hydrogen pops with a splint, oxygen relights a splint and ammonia turns litmus blue, so the gas that turns limewater milky is carbon dioxide.
    Method:

    Match the limewater test to the gas it identifies.
    Examiner tips

    • Limewater milky = carbon dioxide.

21. Question 3(f)

    1 marksIdentifying the anion from a barium nitrate test
    Step 1: Acidified barium nitrate is the test for sulfate ions. Step 2: Sulfate ions give a white precipitate of barium sulfate, which is insoluble. Step 3: Chloride is tested with silver nitrate, carbonate gives carbon dioxide with acid and nitrate gives no precipitate here, so the anion is sulfate.
    Method:

    Match the white precipitate with barium nitrate to the sulfate ion.
    Examiner tips

    • Acidified barium nitrate + sulfate -> white precipitate of barium sulfate.

22. Question 4

    6 marksPlanning a reactivity investigation by temperature rise
    Step 1: To compare fairly, use the same fixed volume of copper(II) sulfate and the same mass of metal each time, measure the initial temperature, add the metal and stir, then record the highest temperature reached. Step 2: The temperature rise (highest minus initial) is found for each metal under identical conditions, so the only variable is the metal. Step 3: The bigger the reactivity difference with copper, the larger the temperature rise, so the metal giving the biggest rise is the most reactive and the smallest rise the least reactive, giving the order of reactivity.
    Method:

    Design a fair test that varies only the metal, then rank the metals by their temperature rise.
    Examiner tips

    • Fair test: same volume and mass; compare the temperature rise for each metal.