IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 12: Experimental Techniques and Chemical Analysis -- Mock Exam 1

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

Question 1 -- Apparatus and Gas Collection

Total: 12 marks

A student in Pune prepares gases in the school laboratory and must choose suitable apparatus for measuring and collecting them.

(a) [4]

Name the most suitable piece of apparatus for each task.

(i) measuring exactly 25.0 cm³ of solution for a titration [1]
(ii) adding a variable volume of acid, read to the nearest 0.05 cm³ [1]
(iii) measuring approximately 50 cm³ of water [1]
(iv) measuring the time taken for a precipitate to form [1]

Model Answer -- 1(a)

(i) (volumetric) pipette [1]

(ii) burette [1]

(iii) measuring cylinder [1]

(iv) stopwatch / stop-clock [1]

⚠ If you missed marks here: Match the accuracy of the apparatus to the wording of the task: a fixed accurate volume means pipette, a variable accurate volume means burette, and the word "approximately" signals a measuring cylinder. Writing "pipette" for the variable volume or "burette" for the fixed volume is the classic swap that loses two marks at once.

1 mark -- volumetric pipette for fixed 25.0 cm3 1 mark -- burette for variable volume to 0.05 cm3 1 mark -- measuring cylinder for approximate 50 cm3 1 mark -- stopwatch for time taken

(b) [4]

The student prepares hydrogen, which is insoluble in water and much less dense than air.

(i) Name two different methods that could be used to collect the hydrogen. [2]

(ii) For each method, give the property of hydrogen that makes it suitable. [2]

Model Answer -- 1(b)

(i) collection over water [1]

(i) upward delivery (into an inverted container) [1]

(ii) over water is suitable because hydrogen is insoluble in water [1]

(ii) upward delivery is suitable because hydrogen is less dense than air [1]

⚠ If you missed marks here: Each collection method must be justified by the matching property: solubility decides whether collection over water is possible, and density compared with air decides the direction of delivery. Pairing "over water" with "less dense than air" mixes the two justifications and loses the explanation marks even though both facts are individually true.

1 mark -- collection over water named 1 mark -- upward delivery named 1 mark -- over water works because hydrogen insoluble 1 mark -- upward delivery works because hydrogen less dense than air

The student needs a dry sample of carbon dioxide, which is denser than air and slightly soluble in water.

(i) Explain why collection over water is a poor choice for this sample. [2]

(ii) Describe how a dry sample of carbon dioxide should be obtained and collected. [2]

Model Answer -- 1(c)

(i) some of the carbon dioxide dissolves in the water, so gas is lost [1]

(i) the collected gas is wet / contaminated with water vapour [1]

(ii) pass the gas through a drying agent, e.g. concentrated sulfuric acid or anhydrous calcium chloride [1]

(ii) collect by downward delivery into an upright container because carbon dioxide is denser than air [1]

⚠ If you missed marks here: Two separate problems with collection over water must be stated: loss of gas by dissolving, and the sample becoming wet — one point alone earns only one mark. In (ii) the drying step must come before collection, and the collection method must be a dry one; drying the gas and then bubbling it through water again is a contradiction examiners look out for.

1 mark -- gas partly dissolves so volume lost 1 mark -- collected gas contaminated with water vapour 1 mark -- dry gas with concentrated sulfuric acid or anhydrous calcium chloride 1 mark -- collect by downward delivery because denser than air

Question 2 -- Acid-Base Titration

Total: 12 marks

A student in York determines the concentration of dilute hydrochloric acid by titrating it against 25.0 cm³ portions of 0.150 mol/dm³ aqueous sodium hydroxide.

(a) [3]

Describe how the student should carry out one accurate titration. Include the names of the apparatus used and the use of an indicator. You may assume the burette has already been filled with the acid.

Model Answer -- 2(a)

use a (volumetric) pipette to transfer 25.0 cm³ of sodium hydroxide into a conical flask, and add a few drops of indicator such as methyl orange or thymolphthalein [1]

add acid from the burette with swirling of the flask [1]

add the acid drop by drop near the end-point and stop when the indicator just changes colour; read the burette to the nearest 0.05 cm³ [1]

⚠ If you missed marks here: The three ingredients of a full-mark answer are the pipette with indicator, continuous swirling, and dropwise addition at the end-point. Vague phrases such as "add the acid slowly" do not earn the dropwise mark, and forgetting the indicator entirely makes the whole titration impossible to follow.

1 mark -- pipette 25.0 cm3 alkali into conical flask with few drops of indicator 1 mark -- add acid from burette while swirling flask 1 mark -- dropwise near end-point until indicator just changes colour

(b) [4]

The student’s four titres are shown.

titration	1 (rough)	2	3	4
titre / cm³	24.90	24.30	24.25	24.35

(i) State which titres should be used to calculate the mean, and explain why. [2]

(ii) Calculate the mean titre. [1]

(iii) Explain why the result of titration 1 is not included. [1]

Model Answer -- 2(b)

(i) use titrations 2, 3 and 4 [1]

(i) because they are concordant / within 0.10 cm³ of each other [1]

(ii) mean = (24.30 + 24.25 + 24.35) ÷ 3 = 24.30 cm³ [1]

(iii) titration 1 is the rough titration and is more than 0.10 cm³ from the others, so it is not concordant [1]

⚠ If you missed marks here: The examiner wants the word "concordant" or the phrase "within 0.10 cm³" — saying the titres are "close together" is not precise enough. Including the rough titre in the mean shifts the answer to 24.45 cm³ and loses the calculation mark as well, so the selection step must come before the arithmetic.

1 mark -- titrations 2 3 and 4 selected 1 mark -- concordant within 0.10 cm3 stated 1 mark -- mean titre 24.30 cm3 1 mark -- rough titre excluded as not concordant

The equation for the reaction is HCl + NaOH → NaCl + H₂O.

Using the mean titre of 24.30 cm³, calculate the concentration of the hydrochloric acid in mol/dm³. Give your answer to three significant figures.

(i) Calculate the number of moles of NaOH in 25.0 cm³ of 0.150 mol/dm³ solution. [1]
(ii) State the number of moles of HCl that reacted. [2]
(iii) Calculate the concentration of the hydrochloric acid. [2]

Model Answer -- 2(c)

(i) moles NaOH = 0.0250 × 0.150 = 0.00375 mol [1]

(ii) the equation shows a 1 : 1 ratio [1]

(ii) so moles HCl = 0.00375 mol [1]

(iii) concentration = 0.00375 ÷ 0.02430 [1]

(iii) = 0.154 mol/dm³ (3 significant figures) [1]

⚠ If you missed marks here: Every titration calculation follows the same three steps: moles of the known solution, mole ratio from the equation, then concentration = moles ÷ volume in dm³. The commonest error is dividing by 24.30 instead of 0.02430 — the titre must be converted from cm³ to dm³ before the final division, and the answer must be rounded to the stated number of significant figures.

1 mark -- moles NaOH 0.00375 1 mark -- one to one mole ratio stated 1 mark -- moles HCl 0.00375 1 mark -- divide by 0.02430 dm3 1 mark -- answer 0.154 mol per dm3 to 3 sig figs

Question 3 -- Chromatography of Food Dyes

Total: 12 marks

A food inspector in Manchester uses paper chromatography to check whether a sweet contains only permitted colourings.

(a) [4]

Describe how the inspector should set up and run the chromatography experiment, starting with a strip of chromatography paper and a solution of the sweet’s colouring.

Model Answer -- 3(a)

draw a baseline in pencil near the bottom of the paper and place a small spot of the colouring on it [1]

stand the paper in a beaker of solvent with the solvent level below the baseline [1]

cover the beaker and leave until the solvent has risen near the top of the paper [1]

remove the paper and immediately mark the position of the solvent front [1]

⚠ If you missed marks here: Two positional details carry marks that are easily thrown away: the baseline must be in pencil, and the solvent must start below the baseline. Marking the solvent front the moment the paper comes out is the step most students forget — once the paper dries, the front is invisible and no R_f value can be calculated.

1 mark -- pencil baseline with small spot of colouring 1 mark -- solvent level below baseline 1 mark -- covered beaker solvent rises near top 1 mark -- mark solvent front immediately on removal

(b) [4]

On the chromatogram, one dye spot travels 4.5 cm from the baseline while the solvent front travels 7.5 cm.

(i) Write the equation used to calculate an R_f value. [1]
(ii) Calculate the R_f value of this dye. [1]
(iii) Explain why the baseline is drawn in pencil and not ink. [1]
(iv) Explain why R_f values are always less than 1. [1]

Model Answer -- 3(b)

(i) R_f = distance moved by substance ÷ distance moved by solvent [1]

(ii) R_f = 4.5 ÷ 7.5 = 0.60 [1]

(iii) ink would dissolve in the solvent and move up the paper, contaminating the chromatogram; pencil is insoluble [1]

(iv) a dissolved substance is carried by the solvent so it can never travel further than the solvent front [1]

⚠ If you missed marks here: The R_f fraction must be the right way up — substance distance over solvent distance — and an answer greater than 1 should ring an alarm bell immediately. R_f values have no units and are quoted as decimals; writing 0.60 cm loses the mark because the units cancel in the ratio.

1 mark -- Rf equation substance distance over solvent distance 1 mark -- Rf value 0.60 no units 1 mark -- ink dissolves and runs pencil insoluble 1 mark -- substance cannot overtake solvent front

The chromatogram compares the sweet’s colouring, S, with three permitted dyes D1, D2 and D3. S gives three spots: two level with D1 and D3, and one that matches none of the permitted dyes.

(i) What does the chromatogram show about whether S is pure? Explain your answer. [2]

(ii) What conclusions can be drawn about the composition of S? [2]

Model Answer -- 3(c)

(i) S is not pure / is a mixture [1]

(i) because it produces more than one spot; a pure substance gives a single spot [1]

(ii) S contains the permitted dyes D1 and D3, shown by spots at matching heights / matching R_f values [1]

(ii) S also contains an unidentified dye that is not one of the permitted colourings, so the sweet fails the check [1]

⚠ If you missed marks here: Conclusions must be tied to the evidence: spots at the same height mean the same R_f value and therefore the same dye. The unmatched spot is the key finding — an answer that only lists D1 and D3 and ignores the third spot misses the whole point of the inspection.

1 mark -- S not pure it is a mixture 1 mark -- more than one spot pure gives single spot 1 mark -- contains D1 and D3 matching spot heights 1 mark -- extra spot is unidentified non-permitted dye

Question 4 -- Separation and Purification

Total: 12 marks

Rock salt from a mine in Cheshire is a mixture of sodium chloride and insoluble sand and grit.

(a) [4]

Describe how pure, dry crystals of sodium chloride can be obtained from crushed rock salt. Use the terms residue and filtrate in your answer.

Model Answer -- 4(a)

add water and stir to dissolve the sodium chloride [1]

filter: the sand and grit remain on the paper as the residue [1]

the salt solution passes through as the filtrate; heat it to the point of crystallisation and allow to cool so crystals form [1]

filter off the crystals and dry them between filter papers / in a warm oven [1]

⚠ If you missed marks here: The question explicitly asks for the words residue and filtrate, so an answer that never uses them cannot earn full marks — the sand is the residue and the salt solution is the filtrate. The final drying step is worth a mark of its own and is the one most often left out.

1 mark -- dissolve salt in water by stirring 1 mark -- filter with sand as residue 1 mark -- filtrate heated to crystallisation point then cooled 1 mark -- crystals dried between filter papers

(b) [4]

Choose and justify a separation method for each task.

(i) obtaining pure water from sea water [2]

(ii) obtaining ethanol (boiling point 78 °C) from a mixture of ethanol and water [2]

Model Answer -- 4(b)

(i) simple distillation [1]

(i) the water boils off and is condensed and collected, leaving the dissolved salts behind [1]

(ii) fractional distillation [1]

(ii) the two miscible liquids have different boiling points, and the fractionating column allows the more volatile ethanol to be collected first [1]

⚠ If you missed marks here: The distinction is what is being separated: a liquid from dissolved solids needs only simple distillation, but two miscible liquids need a fractionating column. Justifications must mention what happens physically — naming the method without saying why it works earns only half the marks.

1 mark -- simple distillation for sea water 1 mark -- water evaporated condensed salts left behind 1 mark -- fractional distillation for ethanol and water 1 mark -- different boiling points column separates miscible liquids

A batch of paracetamol made in a factory should melt sharply at 169 °C. A sample from the batch melts between 158 °C and 165 °C.

(i) What do these results show about the sample? Explain your answer. [2]

(ii) Explain why it matters that a drug is pure before it is sold. [2]

Model Answer -- 4(c)

(i) the sample is impure [1]

(i) a pure substance melts sharply at one temperature; impurities lower the melting point and spread it over a range [1]

(ii) impurities in a drug may be harmful / toxic to the patient [1]

(ii) impurities also change the amount of active drug present, so the dose would be wrong [1]

⚠ If you missed marks here: Both halves of the evidence matter: the melting point is low AND spread over a range, and each observation points to impurity. In (ii) two separate consequences are needed for two marks — harm from the impurity itself and an incorrect dose of the active ingredient are different points, not one point said twice.

1 mark -- sample impure 1 mark -- pure melts sharply impurities lower and broaden melting point 1 mark -- impurities may be toxic to patient 1 mark -- amount of active drug wrong so dose incorrect

Question 5 -- Identifying Cations

Total: 10 marks

A technician in Dubai finds four unlabelled bottles of colourless or pale solutions. Each contains one of: copper(II) sulfate, iron(III) chloride, ammonium chloride and zinc sulfate.

(a) [4]

Aqueous sodium hydroxide is added slowly to a sample of each solution. State what is observed with:

(i) the copper(II) sulfate solution [1]
(ii) the iron(III) chloride solution [1]
(iii) the ammonium chloride solution, on warming [1]
(iv) the zinc sulfate solution, when excess sodium hydroxide is added [1]

Model Answer -- 5(a)

(i) a light blue precipitate forms (insoluble in excess) [1]

(ii) a red-brown precipitate forms [1]

(iii) a gas (ammonia) is given off which turns damp red litmus paper blue [1]

(iv) a white precipitate forms which dissolves in excess to give a colourless solution [1]

⚠ If you missed marks here: Observation questions want colours and behaviour in excess, not ion names: "light blue precipitate" scores, "copper hydroxide forms" does not. For ammonium the observation is the gas and its litmus result, and for zinc the crucial detail is that the white precipitate redissolves in excess alkali.

1 mark -- light blue precipitate for copper 1 mark -- red-brown precipitate for iron three 1 mark -- ammonia gas turns damp red litmus blue 1 mark -- white precipitate dissolves in excess for zinc

(b) [3]

A fifth solution contains either aluminium ions or zinc ions. Both give a white precipitate with sodium hydroxide that dissolves in excess. Describe how aqueous ammonia can be used to find out which ion is present, and give the result for each ion.

Model Answer -- 5(b)

add aqueous ammonia slowly and then in excess: both ions first give a white precipitate [1]

if the ion is zinc, the precipitate dissolves in excess ammonia [1]

if the ion is aluminium, the precipitate does not dissolve in excess ammonia [1]

⚠ If you missed marks here: Repeating the sodium hydroxide test cannot separate these two ions because they behave identically with it — the ammonia test is the only distinguishing test on the syllabus. Both results must be stated: what zinc does AND what aluminium does, because a distinguishing test needs a different outcome for each candidate.

1 mark -- add ammonia slowly then excess white precipitate both 1 mark -- zinc precipitate dissolves in excess ammonia 1 mark -- aluminium precipitate insoluble in excess ammonia

Flame tests are used on three solid salts.

(i) Describe how a flame test is carried out. [1]

(ii) Give the flame colours produced by a sodium salt and by a potassium salt. [2]

Model Answer -- 5(c)

(i) clean a (nichrome/platinum) wire in concentrated hydrochloric acid, dip it into the solid and hold it in a hot / roaring Bunsen flame [1]

(ii) sodium: yellow flame [1]

(ii) potassium: lilac flame [1]

⚠ If you missed marks here: The procedure mark needs the cleaning step with acid as well as holding the sample in the flame — a dirty wire gives false colours. Learn the syllabus colour words exactly: yellow for sodium and lilac for potassium; "purple" is usually accepted for lilac but "orange" for sodium is not.

1 mark -- clean wire in acid dip in solid hold in flame 1 mark -- sodium yellow flame 1 mark -- potassium lilac flame

Question 6 -- Identifying Salt X

Total: 12 marks

A student in Kochi is given a white crystalline solid, X, and carries out the tests shown. Test 1: a flame test gives a yellow colour. Test 2: dilute hydrochloric acid is added to X; there is no effervescence. Test 3: X is dissolved in water, acidified with dilute nitric acid, and aqueous silver nitrate is added; a white precipitate forms. Test 4: a separate acidified solution of X gives no precipitate with aqueous barium nitrate.

(a) [4]

(i) What does Test 1 show about X? [1]

(ii) What does Test 2 show about X? [1]

(iii) What does Test 3 show about X? [1]

(iv) What does Test 4 show about X? [1]

Model Answer -- 6(a)

(i) the yellow flame shows X contains sodium ions [1]

(ii) no effervescence shows X does not contain carbonate ions [1]

(iii) the white precipitate with acidified silver nitrate shows X contains chloride ions [1]

(iv) no precipitate with acidified barium nitrate shows X does not contain sulfate ions [1]

⚠ If you missed marks here: Each test must be translated into a statement about one ion, and negative results are just as informative as positive ones — no fizzing rules out carbonate and no barium precipitate rules out sulfate. Answers that only interpret the positive tests give an incomplete picture of X.

1 mark -- yellow flame means sodium ions 1 mark -- no fizz means no carbonate 1 mark -- white precipitate with silver nitrate means chloride 1 mark -- no barium precipitate means no sulfate

(b) [4]

(i) Identify salt X. [1]

(ii) In Test 3, explain why the solution is acidified with dilute nitric acid before the silver nitrate is added. [2]

(iii) State the colour of the precipitate that would have formed in Test 3 if X had contained iodide ions instead. [1]

Model Answer -- 6(b)

(i) X is sodium chloride [1]

(ii) the nitric acid reacts with / removes any carbonate ions [1]

(ii) carbonate ions would otherwise give a white precipitate (silver carbonate) and a false positive result [1]

(iii) a yellow precipitate (silver iodide) [1]

⚠ If you missed marks here: The identity must combine both detected ions — sodium and chloride — into one named salt. The acidification explanation needs cause and consequence: the acid destroys carbonates, and without it a carbonate would mimic the chloride result. Halide precipitate colours run white, cream, yellow for chloride, bromide, iodide.

1 mark -- X identified as sodium chloride 1 mark -- nitric acid removes carbonate ions 1 mark -- carbonate would give false white precipitate 1 mark -- iodide would give yellow precipitate

A second salt, Y, is known to be either sodium sulfate or sodium sulfite.

(i) Describe a test, and its results, that would distinguish between these two possibilities. [3]

(ii) State why a flame test cannot distinguish between them. [1]

Model Answer -- 6(c)

(i) add acidified aqueous potassium manganate(VII) to a solution of Y [1]

(i) if Y is sodium sulfite the purple colour is decolourised [1]

(i) if Y is sodium sulfate the purple colour remains [1]

(ii) both salts contain the same metal ion (sodium), so both give an identical yellow flame [1]

⚠ If you missed marks here: A distinguishing test needs the reagent plus the result for each salt — three separate marking points. The flame test question is testing whether you understand that flame colours identify the cation only; two sodium salts are indistinguishable in a flame no matter which anions they carry.

1 mark -- acidified potassium manganate seven added 1 mark -- sulfite decolourises purple solution 1 mark -- sulfate leaves purple colour unchanged 1 mark -- same cation so identical yellow flame

Question 7 -- Testing Gases

Total: 10 marks

Four unlabelled gas jars in a school laboratory in Bristol contain hydrogen, oxygen, carbon dioxide and ammonia.

(a) [4]

For each gas, describe the test and the positive result that identifies it.

(i) hydrogen [1]
(ii) oxygen [1]
(iii) carbon dioxide [1]
(iv) ammonia [1]

Model Answer -- 7(a)

(i) a lighted splint gives a squeaky pop [1]

(ii) a glowing splint relights [1]

(iii) bubbled through limewater, the limewater turns milky [1]

(iv) damp red litmus paper turns blue [1]

⚠ If you missed marks here: The splint state is part of the answer: lighted for hydrogen, glowing for oxygen — swapping them turns both marks to zero. For ammonia the litmus paper must be damp and must start red; dry paper gives no result because the gas needs water to act as an alkali.

1 mark -- hydrogen lighted splint squeaky pop 1 mark -- oxygen relights glowing splint 1 mark -- carbon dioxide turns limewater milky 1 mark -- ammonia turns damp red litmus blue

(b) [3]

A fifth gas jar contains either sulfur dioxide or carbon dioxide. Both gases are colourless and acidic. Describe how acidified aqueous potassium manganate(VII) can be used to find out which gas is present, and give the result for each gas.

Model Answer -- 7(b)

bubble the gas through acidified aqueous potassium manganate(VII) [1]

sulfur dioxide turns the solution from purple to colourless [1]

carbon dioxide leaves the purple colour unchanged [1]

⚠ If you missed marks here: Because both gases are acidic, litmus cannot separate them — the manganate(VII) test works because sulfur dioxide is a reducing agent and carbon dioxide is not. Both outcomes must be stated for full marks, and the colour change direction is purple to colourless, never the reverse.

1 mark -- bubble gas through acidified potassium manganate seven 1 mark -- sulfur dioxide purple to colourless 1 mark -- carbon dioxide no colour change

Chlorine is produced at the anode during the electrolysis of concentrated aqueous sodium chloride.

(i) Describe the test for chlorine and its result. [2]

(ii) Suggest one safety precaution that should be taken when testing chlorine, and give a reason. [1]

Model Answer -- 7(c)

(i) hold damp (blue) litmus paper in the gas [1]

(i) the litmus paper is bleached / turns white (it may turn red first) [1]

(ii) carry out the test in a fume cupboard / use only a small amount, because chlorine is toxic [1]

⚠ If you missed marks here: The key word for the chlorine result is bleaches — "changes colour" or "turns red" alone is not enough, although mentioning the brief red stage first is fine. The safety mark needs both the precaution and the reason; "be careful" with no reason scores nothing.

1 mark -- damp litmus paper held in gas 1 mark -- paper bleached turns white 1 mark -- fume cupboard because chlorine toxic

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