IGCSE Chemistry 0620 Paper 5 Practice Paper (Practical Test) — 40 marks, 1 hour 15 minutes, answers and marking scheme

2026-10-30

IGCSE Chemistry practice – Core

An original practice paper written to the Cambridge IGCSE Chemistry 0620 Paper 5 format: Core level, 4 structured questions, 20 parts, 40 marks and 1 hour 15 minutes. All questions are original and are not past paper questions or an official mark scheme.

Total 40 marks · 1 hour 15 minutes · 20 parts

Mark distribution 14 + 13 + 14 + 14 + 13 + 12 = 80

Level Core (same content as Paper 1 and Paper 5)

Paper at a glance

Paper	Level	Time	Marks
Paper 5	Core	1 hour 15 minutes	40 marks

Questions

Total: 40 marks · 1 hour 15 minutes · calculator allowed

Question 1 · Acid-base titration and calculation [10]

(a) State the apparatus used to measure exactly 25.0 cm3 of sodium hydroxide solution for a titration. [1]

(b) Explain why the burette is rinsed with a little of the sulfuric acid before it is filled. [2]

(c) Four titres are 12.4, 13.9, 12.5 and 12.6 cm3. Identify the anomalous titre and calculate the mean of the other three. [2]

(d) 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide is neutralised by a mean titre of 12.5 cm3 of sulfuric acid (H2SO4 + 2NaOH -> Na2SO4 + 2H2O). Calculate the concentration of the acid. Show your working. [3]

(e) Name the indicator used in this titration and state the colour change at the end point. [2]

Question 2 · Calorimetry and enthalpy of neutralisation [10]

(a) Explain why the solutions are mixed in a polystyrene cup and why the cup is covered after mixing. [2]

(b) After mixing, the temperature rises from 25.0 °C to 28.4 °C. The total mass of solution is 100 g and its specific heat capacity is 4.2 J/(g °C). Calculate the energy released. Show your working. [3]

(c) 50.0 cm3 of 0.500 mol/dm3 hydrochloric acid is exactly neutralised by 50.0 cm3 of 0.500 mol/dm3 sodium hydroxide. Use your answer to (b) to calculate the enthalpy change of neutralisation in kJ/mol, with the correct sign. [3]

(d) The literature value is -57.3 kJ/mol. Give one reason why the experimental value of -57.1 kJ/mol is slightly different. [1]

(e) Suggest one improvement that would reduce heat loss in this experiment. [1]

Question 3 · Rate of reaction and gas collection [10]

(a) Magnesium reacts with dilute hydrochloric acid and the hydrogen is collected in a gas syringe. In the first 20 s, 30 cm3 of hydrogen is collected. Calculate the initial rate in cm3/s. [2]

(b) After 60 s a total of 72 cm3 of hydrogen has been collected. Calculate the average rate from 0 to 60 s. [2]

(c) At room temperature the molar volume of a gas is 24 dm3/mol. Use the 72 cm3 of hydrogen to calculate the mass of magnesium that has reacted (Mg = 24). [3]

(d) Explain why the slope of the volume-time graph, and therefore the rate of reaction, gradually decreases between 0 and 60 s. [2]

(e) To compare the effect of temperature on the rate, which condition must be kept constant apart from temperature? [1]

Question 4 · Tests for ions and chromatography [10]

(a) An unknown solution gives a lilac colour in a flame test. Name the cation it contains and explain why the flame is viewed through blue cobalt glass. [2]

(b) Adding sodium hydroxide solution to an unknown solution gives a white precipitate. The precipitate dissolves when excess sodium hydroxide is added. Name one possible cation. [2]

(c) Describe the test and the observation that would confirm the presence of sulfate ions in a solution. [2]

(d) In a chromatogram a spot moves 4.8 cm from the start line while the solvent front moves 6.0 cm. Calculate the Rf value of the spot. [2]

(e) State why the start line is drawn in pencil rather than in ink, and give one experimental condition that affects the Rf value. [2]

Answers and marking scheme

Part	Answer	Marking points	Marks
1(a)	A 25.0 cm3 volumetric pipette. A measuring cylinder is not accurate enough for a titration.	1 mark: pipette / volumetric pipette.	1
1(b)	Rinsing removes any distilled water left in the burette. Otherwise the acid would be diluted, a larger titre would be measured and the calculated concentration would be inaccurate.	1 mark: removes leftover water / prevents dilution of the acid. 1 mark: keeps the titre and the calculated concentration accurate.	2
1(c)	13.9 cm3 is clearly further from the others, so it is anomalous. Mean = (12.4 + 12.5 + 12.6) / 3 = 12.5 cm3.	1 mark: 13.9 cm3 identified as anomalous. 1 mark: 12.5 cm3.	2
1(d)	Moles of NaOH = 25.0 / 1000 x 0.100 = 2.50 x 10-3 mol. The equation ratio is 1:2, so moles of H2SO4 = 2.50 x 10-3 / 2 = 1.25 x 10-3 mol. Concentration = 1.25 x 10-3 / (12.5 / 1000) = 0.100 mol/dm3.	1 mark: moles of NaOH = 2.50 x 10-3 mol. 1 mark: divided by 2 gives 1.25 x 10-3 mol of H2SO4. 1 mark: 0.100 mol/dm3 (with unit).	3
1(e)	Phenolphthalein. The solution is pink in alkali and the pink colour just disappears (turns colourless) at the end point.	1 mark: phenolphthalein. 1 mark: pink -> colourless.	2
2(a)	Polystyrene is a poor conductor of heat, so less heat is lost through the cup. The lid reduces heat loss by evaporation and convection, so the measured temperature rise is closer to the heat actually released.	1 mark: polystyrene insulates / reduces heat loss. 1 mark: lid reduces loss by evaporation and convection.	2
2(b)	Temperature rise = 28.4 – 25.0 = 3.4 °C. q = m c dT = 100 x 4.2 x 3.4 = 1428 J (about 1.43 kJ).	1 mark: dT = 3.4 °C. 1 mark: uses q = m c dT. 1 mark: 1428 J (or 1.43 kJ).	3
2(c)	Moles of water formed = 50.0 / 1000 x 0.500 = 0.0250 mol. dH = -1428 J / 0.0250 mol = -57 120 J/mol = about -57.1 kJ/mol. The sign is negative because heat is released.	1 mark: 0.0250 mol of water formed. 1 mark: -1428 / 0.0250 = -57 120 J/mol. 1 mark: -57.1 kJ/mol (sign and unit).	3
2(d)	Some heat is lost to the surroundings during mixing and temperature measurement because the cup is not perfectly insulated, so the measured temperature rise and the calculated energy are both too small.	1 mark: heat lost to the surroundings / not perfectly insulated.	1
2(e)	Wrap the polystyrene cup in insulating material such as cloth or cotton wool, replace the lid at once after mixing and read the highest temperature as quickly as possible.	1 mark: add insulation / replace the lid at once / read the highest temperature quickly.	1
3(a)	Initial rate = 30 cm3 / 20 s = 1.5 cm3/s.	1 mark: 30 / 20. 1 mark: 1.5 cm3/s (with unit).	2
3(b)	Average rate = 72 cm3 / 60 s = 1.2 cm3/s.	1 mark: 72 / 60. 1 mark: 1.2 cm3/s.	2
3(c)	Moles of H2 = 72 / 24 000 = 3.0 x 10-3 mol. From Mg + 2HCl -> MgCl2 + H2 the ratio is 1:1, so moles of Mg = 3.0 x 10-3 mol. Mass = 3.0 x 10-3 x 24 = 0.072 g.	1 mark: 3.0 x 10-3 mol of H2. 1 mark: 1:1 ratio, same moles of Mg. 1 mark: 0.072 g.	3
3(d)	The concentration of hydrogen ions in the acid falls as they are used up, so particles collide less frequently and the rate falls. The surface area of the magnesium also decreases as it is used up.	1 mark: acid concentration falls / fewer collisions. 1 mark: magnesium surface area decreases.	2
3(e)	The mass and surface area of the magnesium, with the same concentration and volume of acid.	1 mark: mass / surface area of magnesium (or acid concentration and volume) kept constant.	1
4(a)	Potassium ion, K+. Any sodium present gives an intense yellow flame that hides the lilac, so blue cobalt glass is used to filter out the yellow light and show the lilac of K+ clearly.	1 mark: K+ (potassium ion). 1 mark: cobalt glass filters out the yellow sodium light / stops it hiding the lilac.	2
4(b)	Zinc ion, Zn2+ (or aluminium ion, Al3+), because both give a white precipitate with a little sodium hydroxide that redissolves in excess sodium hydroxide.	1 mark: Zn2+ or Al3+. 1 mark: white precipitate soluble in excess NaOH.	2
4(c)	Acidify with dilute hydrochloric acid, then add barium chloride solution. A white precipitate of barium sulfate confirms sulfate ions, SO4 2-.	1 mark: add dilute hydrochloric acid, then barium chloride solution. 1 mark: white precipitate.	2
4(d)	Rf = distance moved by the spot / distance moved by the solvent front = 4.8 / 6.0 = 0.80.	1 mark: 4.8 / 6.0. 1 mark: 0.80.	2
4(e)	Pencil lead is graphite, which does not dissolve in the solvent, so it does not move up and interfere with the spots. The Rf value depends on the solvent used, the temperature and the type of paper.	1 mark: pencil (graphite) is insoluble in the solvent. 1 mark: solvent used / temperature / type of paper.	2

Timing plan

Section	Suggested time	Marks
Questions 1-2	25 min	27 marks
Questions 3-4	26 min	28 marks
Questions 5-6	24 min	25 marks
Checking	remaining time	–

Common pitfalls

Calculations written without working lose every method mark.
Explanations that list key words without a cause-and-effect sentence.
Missing units: cm3 s-1, g h-1, dm3 min-1, plants m-2.
Core answers may be shorter, but you still need one point per mark available.

How to use this paper Sit the paper closed-book in 75 minutes, then mark point by point with the answer booklet and redo every lost mark.

Syllabus reference Format follows Cambridge IGCSE Chemistry 0620 (2026-2028) Paper 5 Core: 1 hour 15 minutes, 40 marks.

Frequently asked questions

Who is this paper for?
Students preparing for IGCSE Chemistry 0620 Core papers (Paper 1 and Paper 5), especially those who need practice in writing structured answers.

Does the answer booklet include a mark scheme?
Yes. Every part lists its marking points so you can mark yourself or a tutor can mark point by point.

Can I print it?
Yes, both the question paper and the answer booklet are A4 PDFs ready to print.

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Originality statement: every question, data set and context in this paper was written by TutorZone. It is not a Cambridge IGCSE Chemistry 0620 past paper or an official mark scheme.

This paper is not an IGCSE Chemistry 0620 past paper; only the structure and mark distribution follow the official Paper 5 (Core) format.