IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 10: Chemistry of the Environment -- Challenge Paper

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 -- Comparing Nitrogen Fertilisers

Total: 12 marks

An agricultural co-operative in Gujarat must choose between two nitrogen fertilisers for a 40 hectare wheat crop. Ammonium nitrate, NH₄NO₃, costs $420 per tonne and urea, CO(NH₂)₂, costs $530 per tonne. The crop requires 120 kg of nitrogen per hectare.

(a) [4]

(i) Calculate the percentage by mass of nitrogen in ammonium nitrate and in urea. (Aᵣ: H = 1, C = 12, N = 14, O = 16) [3]

(ii) State which compound is the richer source of nitrogen per kilogram of fertiliser. [1]

Model Answer -- 1(a)

(i) Mᵣ of NH₄NO₃ = 80 and Mᵣ of CO(NH₂)₂ = 60 [1]

(i) Percentage of nitrogen in ammonium nitrate = 28 ÷ 80 × 100 = 35.0% [1]

(i) Percentage of nitrogen in urea = 28 ÷ 60 × 100 = 46.7% [1]

(ii) Urea, because it contains a higher percentage of nitrogen by mass [1]

⚠ If you missed marks here: Both compounds contain exactly two nitrogen atoms, so the only thing that differs is the formula mass; a smaller Mᵣ therefore means a richer fertiliser. Expanding CO(NH₂)₂ carelessly to give Mᵣ = 46 or 76 is the usual error, and it wrecks every later part of this question, so check that expansion before going on.

1 mark -- formula masses 80 and 60 1 mark -- ammonium nitrate 35.0 percent nitrogen 1 mark -- urea 46.7 percent nitrogen 1 mark -- urea is the richer source per kilogram

(b) [4]

(i) Calculate the total mass of nitrogen, in kg, needed for the whole 40 hectare crop. [1]

(ii) Calculate the mass of urea, in tonnes, that would supply this nitrogen. [2]

(iii) Calculate the cost of this mass of urea. [1]

Model Answer -- 1(b)

(i) Total nitrogen = 120 × 40 = 4800 kg [1]

(ii) Mass of urea = 4800 ÷ 0.467 = 10 280 kg (method: divide by the fraction of nitrogen) [1]

(ii) This is approximately 10.3 tonnes [1]

(iii) Cost = 10.3 × 530 = $5459, that is about $5460 [1]

⚠ If you missed marks here: The direction of the percentage step decides everything: you need more fertiliser than nitrogen, so divide by 0.467 rather than multiplying by it. Multiplying gives about 2240 kg, an answer far too small to be sensible. Convert to tonnes only at the end, and keep the unit with every line of working.

1 mark -- total nitrogen 4800 kg 1 mark -- method divide 4800 by 0.467 1 mark -- mass of urea about 10.3 tonnes 1 mark -- cost about 5460 dollars

Using ammonium nitrate instead would require 13.7 tonnes, costing $5754.

(i) Deduce which fertiliser is the cheaper source of the nitrogen the crop needs, showing how you decide. [2]

(ii) Suggest two factors other than cost that the co-operative should consider before choosing. [2]

Model Answer -- 1(c)

(i) The two totals must be compared for the same amount of nitrogen delivered: $5460 for urea against $5754 for ammonium nitrate [1]

(i) Urea is cheaper, even though its price per tonne is higher, because far less of it is needed [1]

(ii) Any valid factor, for example: the cost of transporting and storing the larger mass of ammonium nitrate [1]

(ii) A second valid factor, for example: the risk of nitrogen being lost as ammonia gas, the solubility and speed of action of each salt, or the risk of nitrate run-off polluting waterways [1]

⚠ If you missed marks here: Comparing prices per tonne rather than per unit of nitrogen gives exactly the wrong answer, which is the trap this question sets. State clearly that the comparison is made for the same mass of nitrogen supplied. In (ii) the two factors must be genuinely different considerations, not two ways of saying "it is easier to use".

1 mark -- compares total costs for same nitrogen delivered 1 mark -- urea cheaper despite higher price per tonne 1 mark -- first valid non-cost factor such as transport and storage 1 mark -- second valid factor such as nitrogen loss or run-off risk

Question 2 -- Flue Gas Desulfurisation

Total: 12 marks

A power station burns 5000 tonnes of coal each day. The coal contains 1.6% sulfur by mass. All the sulfur is converted to sulfur dioxide, which is removed using calcium carbonate: CaCO₃ + SO₂ → CaSO₃ + CO₂

(a) [4]

(i) Calculate the mass of sulfur burned each day. [1]

(ii) Calculate the mass of sulfur dioxide produced each day. (Aᵣ: O = 16, S = 32) [2]

(iii) Explain why the mass of sulfur dioxide is greater than the mass of sulfur burned. [1]

Model Answer -- 2(a)

(i) Mass of sulfur = 1.6/100 × 5000 = 80 tonnes [1]

(ii) Every 32 tonnes of sulfur gives 64 tonnes of SO₂, so the ratio of masses is 64 : 32 [1]

(ii) Mass of SO₂ = 80 × 64/32 = 160 tonnes [1]

(iii) Each sulfur atom combines with two oxygen atoms from the air, and the mass of that oxygen is added to the mass of the sulfur [1]

⚠ If you missed marks here: The mass ratio S : SO₂ is 32 : 64, so the mass exactly doubles — using 32 : 48 or forgetting the ratio altogether are the usual slips. In (iii) the explanation must mention that the extra mass is oxygen taken from the air, since mass is conserved overall.

1 mark -- 80 tonnes of sulfur burned 1 mark -- mass ratio 64 to 32 identified 1 mark -- 160 tonnes of sulfur dioxide 1 mark -- extra mass is oxygen from the air added to sulfur

(b) [4]

(i) Calculate the mass of calcium carbonate needed each day to remove all this sulfur dioxide. (Mᵣ: CaCO₃ = 100, SO₂ = 64) [3]

(ii) The plant only removes 92% of the sulfur dioxide. Calculate the mass of sulfur dioxide released into the atmosphere each day. [1]

Model Answer -- 2(b)

(i) Amount of SO₂ = 160/64 = 2.5 tonne-moles [1]

(i) The equation shows a 1 : 1 mole ratio, so 2.5 tonne-moles of CaCO₃ are required [1]

(i) Mass of CaCO₃ = 2.5 × 100 = 250 tonnes [1]

(ii) Mass released = 8/100 × 160 = 12.8 tonnes [1]

⚠ If you missed marks here: Reacting-mass calculations must go through moles — assuming equal masses react gives 160 tonnes of limestone instead of 250. Working in tonne-moles is legitimate provided every mass in the calculation is in tonnes. In (ii) remember that the percentage released is 8%, not 92%.

1 mark -- amount of SO2 is 2.5 tonne-moles 1 mark -- one to one mole ratio applied 1 mark -- 250 tonnes of calcium carbonate 1 mark -- 12.8 tonnes of sulfur dioxide released

(i) The desulfurisation process releases carbon dioxide. Explain why this is a disadvantage of the method, and suggest why the operators accept it. [2]

(ii) An engineer suggests using calcium oxide instead of calcium carbonate. Write the equation for the reaction of calcium oxide with sulfur dioxide, and state one advantage of using calcium oxide. [2]

Model Answer -- 2(c)

(i) Carbon dioxide is a greenhouse gas, so the process reduces acid rain but adds to climate change [1]

(i) The operators accept this because the sulfur dioxide is far more damaging locally, causing acid rain and respiratory illness, and limestone is cheap and plentiful [1]

(ii) CaO + SO₂ → CaSO₃ [1]

(ii) Advantage: no carbon dioxide is released by this reaction, and less mass of solid is needed because Mᵣ of CaO is only 56 [1]

⚠ If you missed marks here: This part rewards recognising a trade-off: one pollution problem is reduced while another is slightly worsened. Note also that calcium oxide is itself made by heating limestone, which releases carbon dioxide elsewhere — a sophisticated answer may say so, but the mark here is for the absence of carbon dioxide at the power station.

1 mark -- carbon dioxide is a greenhouse gas adding to climate change 1 mark -- accepted because sulfur dioxide more damaging locally and limestone cheap 1 mark -- equation CaO plus SO2 gives CaSO3 1 mark -- advantage no carbon dioxide released or smaller mass needed

Question 3 -- Evaluating Air Quality Data

Total: 12 marks

A monitoring station beside a road in a European city records annual mean concentrations, in micrograms per cubic metre. Catalytic converters became compulsory on all new cars in 1993, and low-sulfur fuel was introduced in 2005.

(a) [4]

(i) Calculate the percentage decrease in the carbon monoxide concentration between 1990 and 2020. [2]

(ii) Explain, using a balanced equation, how catalytic converters brought about this decrease. [2]

Model Answer -- 3(a)

(i) Decrease = 3200 − 380 = 2820 µg per m³ [1]

(i) Percentage decrease = 2820 ÷ 3200 × 100 = 88.1% [1]

(ii) 2CO + 2NO → 2CO₂ + N₂ [1]

(ii) The catalyst allows carbon monoxide to be oxidised to carbon dioxide inside the exhaust system, so much less escapes into the air [1]

⚠ If you missed marks here: A percentage decrease is always calculated relative to the original value, so the denominator is 3200 and not 380. Dividing the wrong way round gives 742%, an answer that should immediately look impossible for a decrease.

1 mark -- decrease of 2820 micrograms per cubic metre 1 mark -- percentage decrease 88.1 1 mark -- equation 2CO plus 2NO gives 2CO2 plus N2 1 mark -- carbon monoxide oxidised in exhaust so less escapes

(b) [4]

(i) Sulfur dioxide fell by over 90% across the period, while nitrogen dioxide fell by less than half. Explain this difference. [3]

(ii) Traffic increased by 35% over the same period. Explain why this makes the fall in carbon monoxide more impressive than it first appears. [1]

Model Answer -- 3(b)

(i) Sulfur dioxide comes entirely from sulfur compounds in the fuel, so removing the sulfur at the refinery eliminates the emission at source [1]

(i) Oxides of nitrogen are formed from the nitrogen and oxygen of the air itself at the high temperature inside the engine, so they cannot be removed from the fuel [1]

(i) They can only be reduced by the catalytic converter, which does not convert every molecule, so the fall is smaller [1]

(ii) The concentration per vehicle has fallen even further than the totals suggest, because the same measurements come from 35% more traffic [1]

⚠ If you missed marks here: The whole explanation rests on where each pollutant comes from: sulfur from the fuel, nitrogen from the air. Once that contrast is stated, the difference in how easily each can be controlled follows naturally. In (ii) the reasoning is that a fall achieved despite rising traffic implies an even bigger fall per vehicle.

1 mark -- sulfur dioxide comes from fuel so removing sulfur eliminates it 1 mark -- nitrogen oxides formed from air at high engine temperature 1 mark -- only converters reduce them and not completely 1 mark -- fall achieved despite 35 percent more traffic

(i) A politician claims the data prove that "air pollution has been solved". Evaluate this claim using the data. [2]

(ii) The data do not include carbon dioxide. Explain why carbon dioxide emissions from the same traffic are unlikely to have fallen, and why this matters. [2]

Model Answer -- 3(c)

(i) The claim is too strong: although all three pollutants have fallen substantially, nitrogen dioxide remains at 41 µg per m³ and carbon monoxide is still present [1]

(i) The data also cover only three pollutants at one site, and say nothing about particulates or about pollution elsewhere in the city [1]

(ii) Carbon dioxide is the normal product of complete combustion, so it rises with fuel use; catalytic converters actually convert carbon monoxide into more carbon dioxide [1]

(ii) This matters because carbon dioxide is a greenhouse gas, so local air quality can improve while the contribution to climate change continues [1]

⚠ If you missed marks here: Evaluation means acknowledging what the data do support before explaining their limits — one measuring site, three pollutants, no particulates, and levels still above zero. The sharpest point in (ii) is that a catalytic converter increases carbon dioxide output, so cleaner local air does not mean a smaller contribution to global warming.

1 mark -- claim too strong pollutants still present at significant levels 1 mark -- data cover only three pollutants at one site 1 mark -- carbon dioxide is product of complete combustion converters increase it 1 mark -- matters because carbon dioxide is a greenhouse gas

Question 4 -- Designing an Investigation of Water Samples

Total: 12 marks

A student is given four unlabelled water samples: distilled water, tap water, seawater and water taken downstream of a fertiliser factory. She must design an investigation to identify each sample using only school laboratory equipment.

(a) [4]

Describe an investigation that would allow her to distinguish the distilled water from the other three samples, and explain how her results would identify it. Include the apparatus, the measurements taken and the expected results.

Model Answer -- 4(a)

Heat equal volumes of each sample in turn with a thermometer in the liquid and record the temperature at which each boils [1]

Alternatively (or additionally) evaporate a fixed volume of each sample to dryness in a weighed evaporating basin and reweigh [1]

The distilled water boils at exactly 100 °C at standard atmospheric pressure, while the others boil above 100 °C [1]

The distilled water leaves no residue on evaporation, whereas the others leave solid residues [1]

⚠ If you missed marks here: An investigation answer needs both the procedure and the expected results; describing what you would do without saying what you would see throws away half the marks. Note that chemical water tests are useless here, because all four samples would give identical positive results.

1 mark -- heat equal volumes and record boiling temperature 1 mark -- evaporate fixed volume in weighed basin and reweigh 1 mark -- distilled water boils exactly 100 others higher 1 mark -- distilled water leaves no residue others leave solid

(b) [4]

Her results are: sample 1 boils at 100.0 °C, residue 0.00 g; sample 2 boils at 100.1 °C, residue 0.03 g; sample 3 boils at 101.9 °C, residue 3.50 g; sample 4 boils at 100.4 °C, residue 0.31 g. Each residue came from 100 cm³ of sample.

(i) Identify each of the four samples, giving a reason in each case. [4]

Model Answer -- 4(b)

Sample 1 is the distilled water: it boils at exactly 100 °C and leaves no residue at all [1]

Sample 3 is the seawater: it has by far the largest residue and the highest boiling point, consistent with a high concentration of dissolved salts [1]

Sample 2 is the tap water: it contains only a very small amount of dissolved solid, as expected after treatment [1]

Sample 4 is the water from downstream of the factory: its residue is about ten times that of tap water, consistent with dissolved fertiliser salts, but far less than seawater [1]

⚠ If you missed marks here: Work from the extremes inwards: the sample with no residue and the sample with the huge residue are unambiguous, which leaves only two possibilities for the middle pair. Each identification must quote the number that justifies it — bare names without evidence score at most one mark overall.

1 mark -- sample 1 distilled water exactly 100 and no residue 1 mark -- sample 3 seawater largest residue highest boiling point 1 mark -- sample 2 tap water very small residue 1 mark -- sample 4 factory water intermediate residue from dissolved salts

(i) Explain why the boiling-point measurements alone would not have been sufficient to identify all four samples reliably. [2]

(ii) Suggest a further test the student could carry out on sample 4 to support her identification, and state the expected result. [2]

Model Answer -- 4(c)

(i) The boiling points of samples 1, 2 and 4 differ by only a few tenths of a degree, which is within the uncertainty of a school thermometer [1]

(i) The residue masses differ by a factor of ten, so they separate these samples far more reliably than the temperatures do [1]

(ii) Warm a sample of the residue with aqueous sodium hydroxide and test the gas with damp red litmus paper [1]

(ii) If ammonium salts from the fertiliser are present the paper turns blue, confirming the ammonium ion [1]

⚠ If you missed marks here: The evaluation mark depends on comparing the size of the differences with the precision of the instrument — 0.1 °C is not a difference a school thermometer can be trusted to resolve. In (ii) a confirmatory test must be specific to the suspected contaminant, so the ammonium test is the natural choice for fertiliser run-off.

1 mark -- boiling points differ by less than thermometer uncertainty 1 mark -- residue masses differ tenfold so separate samples reliably 1 mark -- warm residue with sodium hydroxide and test gas 1 mark -- damp red litmus turns blue confirming ammonium ion

Question 5 -- Photosynthesis and the Carbon Balance

Total: 10 marks

A reforestation project in Kerala plants trees on 200 hectares of cleared land. Each hectare of young forest is expected to fix 6.0 tonnes of glucose per year by photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

(a) [4]

(i) Calculate the mass of carbon dioxide removed from the atmosphere per hectare per year. (Mᵣ: CO₂ = 44, C₆H₁₂O₆ = 180) [3]

(ii) Calculate the total mass of carbon dioxide removed by the whole project each year. [1]

Model Answer -- 5(a)

(i) Amount of glucose = 6.0/180 = 0.0333 tonne-moles [1]

(i) The equation shows 6 mol of CO₂ per mol of glucose, so amount of CO₂ = 6 × 0.0333 = 0.200 tonne-moles [1]

(i) Mass of CO₂ = 0.200 × 44 = 8.8 tonnes per hectare per year [1]

(ii) Total = 8.8 × 200 = 1760 tonnes per year [1]

⚠ If you missed marks here: The 6 : 1 mole ratio is the heart of this calculation and is the step most often forgotten — leaving it out gives 1.47 tonnes instead of 8.8. Note the mass of carbon dioxide absorbed exceeds the mass of glucose made, because the glucose keeps only some of the oxygen and the rest is released as O₂.

1 mark -- amount of glucose 0.0333 tonne-moles 1 mark -- six to one mole ratio gives 0.200 tonne-moles CO2 1 mark -- 8.8 tonnes per hectare per year 1 mark -- total 1760 tonnes per year

(b) [3]

A nearby coal power station emits 900 000 tonnes of carbon dioxide each year.

(i) Calculate the percentage of the power station's emissions that the forest absorbs. [2]

(ii) Comment on what your answer suggests about relying on tree planting alone to tackle climate change. [1]

Model Answer -- 5(b)

(i) Percentage = 1760 ÷ 900 000 × 100 [1]

(i) = 0.196%, that is about 0.2% [1]

(ii) Tree planting on this scale offsets only a tiny fraction of the emissions, so reducing emissions at source must also be part of any realistic strategy [1]

⚠ If you missed marks here: Show the division before evaluating it, since the method mark is separate from the answer. In (ii) the comment must follow from the number you calculated: a fraction of one per cent means offsetting cannot substitute for cutting emissions, however valuable the forest is for other reasons.

1 mark -- method 1760 divided by 900000 times 100 1 mark -- answer about 0.2 percent 1 mark -- comment that offsetting alone is insufficient must cut emissions

(i) Explain why the carbon absorbed by the forest is returned to the atmosphere if the timber is later burned. [2]

(ii) State the two conditions needed for the trees to photosynthesise. [1]

Model Answer -- 5(c)

(i) The carbon taken from the air is stored in the compounds that make up the wood [1]

(i) Burning the wood oxidises this carbon back to carbon dioxide, which returns to the atmosphere, so the absorption is only temporary unless the timber is preserved [1]

(ii) Light and chlorophyll [1]

⚠ If you missed marks here: Both stages of the carbon's journey are needed: first that it is locked into the wood, then that combustion oxidises it back to carbon dioxide. The same argument applies if the wood is left to rot, since decomposition also returns the carbon to the air.

1 mark -- carbon stored in compounds making up the wood 1 mark -- burning oxidises carbon back to carbon dioxide 1 mark -- light and chlorophyll stated

Question 6 -- Ammonia, Ammonium Salts and Nitrogen Loss

Total: 12 marks

A researcher investigates how much nitrogen is lost as ammonia when ammonium sulfate fertiliser is spread on soil that has recently been treated with lime, calcium hydroxide.

(a) [4]

(i) Write a balanced symbol equation for the reaction between ammonium sulfate and calcium hydroxide. [2]

(ii) Explain, with reference to this reaction, why farmers are advised not to lime a field at the same time as applying an ammonium fertiliser. [2]

Model Answer -- 6(a)

(i) (NH₄)₂SO₄ + Ca(OH)₂ → CaSO₄ + 2NH₃ + 2H₂O — correct formulae [1]

(i) Equation correctly balanced with 2NH₃ and 2H₂O [1]

(ii) The alkaline lime displaces ammonia gas from the ammonium salt [1]

(ii) The ammonia escapes into the air, so the nitrogen is lost instead of being absorbed by the crop, wasting the fertiliser [1]

⚠ If you missed marks here: Each ammonium sulfate unit supplies two ammonium ions, so two molecules each of ammonia and water appear on the right. In (ii) both the chemical event and its agricultural consequence are credited: ammonia is displaced, and the nitrogen escapes to the air rather than feeding the crop.

1 mark -- correct formulae in the equation 1 mark -- balanced with 2NH3 and 2H2O 1 mark -- alkali displaces ammonia from ammonium salt 1 mark -- ammonia escapes so nitrogen lost and fertiliser wasted

(b) [4]

In the laboratory, 13.2 g of ammonium sulfate is warmed with an excess of calcium hydroxide. (Mᵣ: (NH₄)₂SO₄ = 132, NH₃ = 17)

(i) Calculate the maximum mass of ammonia that could be produced. [3]

(ii) Only 2.72 g of ammonia is actually collected. Calculate the percentage yield. [1]

Model Answer -- 6(b)

(i) Amount of (NH₄)₂SO₄ = 13.2/132 = 0.100 mol [1]

(i) Each formula unit gives 2NH₃, so amount of ammonia = 0.200 mol [1]

(i) Mass of ammonia = 0.200 × 17 = 3.40 g [1]

(ii) Percentage yield = 2.72/3.40 × 100 = 80.0% [1]

⚠ If you missed marks here: The 1 : 2 ratio between the salt and the ammonia is the step that decides the answer; missing it gives 1.70 g. In (ii) the percentage yield is the actual mass divided by the theoretical maximum — dividing the other way round gives an impossible figure above 100%.

1 mark -- 0.100 mol of ammonium sulfate 1 mark -- 0.200 mol of ammonia from one to two ratio 1 mark -- theoretical mass 3.40 g 1 mark -- percentage yield 80.0

(i) Describe how the researcher could confirm that the gas collected is ammonia, and explain why this test works. [2]

(ii) Suggest two reasons why the yield of ammonia in the laboratory was less than the theoretical maximum. [2]

Model Answer -- 6(c)

(i) Hold damp red litmus paper in the gas; it turns blue [1]

(i) The test works because ammonia dissolves in the water on the paper to form an alkaline solution, and it is the only common alkaline gas [1]

(ii) Ammonia is very soluble in water, so some dissolves in the water formed or in the damp apparatus [1]

(ii) Some gas escapes from the apparatus, or the reaction is incomplete because the mixture was not warmed for long enough [1]

⚠ If you missed marks here: The explanation mark in (i) needs the chemistry behind the colour change: ammonia dissolves to give an alkaline solution, which is why red litmus turns blue and why the paper must be damp. In (ii) the reasons must be specific to this gas and apparatus — "human error" or "the equipment was inaccurate" never scores.

1 mark -- damp red litmus paper turns blue 1 mark -- ammonia dissolves forming alkaline solution only common alkaline gas 1 mark -- ammonia very soluble so some dissolves in water present 1 mark -- gas escapes or reaction incomplete

Question 7 -- An Unfamiliar Water Supply Problem

Total: 10 marks

A desert town obtains drinking water by distilling seawater in a solar still. The distilled water tastes flat, is slightly acidic and contains no dissolved minerals. Engineers add a small amount of calcium carbonate and a trace of chlorine before the water enters the supply.

(a) [4]

(i) Explain why the distilled water boils at exactly 100 °C at standard atmospheric pressure while the seawater does not. [2]

(ii) Suggest why the distilled water is slightly acidic even though it contains no dissolved solids. [2]

Model Answer -- 7(a)

(i) Distillation evaporates only the water, leaving the dissolved salts behind, so the distillate is pure water with a fixed boiling point of 100 °C [1]

(i) Seawater is a mixture, and its dissolved salts raise the boiling point above 100 °C [1]

(ii) Carbon dioxide from the air dissolves in the water [1]

(ii) It forms a weak acid (carbonic acid) in solution, giving a pH slightly below 7 [1]

⚠ If you missed marks here: Part (ii) is an unfamiliar application of familiar chemistry: the only substance available to dissolve is a gas from the air, and the gas in air that forms an acid is carbon dioxide. Note that a dissolved gas is not a dissolved solid, so there is no contradiction with the statement in the question.

1 mark -- distillation leaves salts behind giving pure water fixed boiling point 1 mark -- seawater dissolved salts raise boiling point 1 mark -- carbon dioxide from the air dissolves 1 mark -- forms weak carbonic acid giving pH below 7

(b) [3]

(i) Explain why the engineers add calcium carbonate to the distilled water before it enters the supply. [2]

(ii) Explain why chlorine is still added, even though distillation has already removed the microbes. [1]

Model Answer -- 7(b)

(i) The calcium carbonate neutralises the dissolved carbon dioxide, raising the pH so the water is not acidic and does not corrode the pipes [1]

(i) It also restores dissolved minerals, improving the taste of the water [1]

(ii) The residual chlorine kills any microbes that enter the water in the storage tanks or distribution pipes after treatment [1]

⚠ If you missed marks here: Two distinct purposes are credited in (i): neutralising the acidity so the pipes are not corroded, and restoring the minerals that give water its taste. In (ii) the mark is for protection after treatment — the water may be sterile as it leaves the still, but it does not stay sealed all the way to the tap.

1 mark -- neutralises acidity raising pH preventing corrosion of pipes 1 mark -- restores dissolved minerals improving taste 1 mark -- residual chlorine kills microbes entering tanks or pipes

A consultant proposes replacing the solar still with a conventional treatment works taking water from a nearby river that receives farm run-off and sewage.

Evaluate this proposal, referring to the treatment stages that would be needed and to the substances present in the river water.

Model Answer -- 7(c)

The works would need sedimentation and filtration to remove the suspended solids, carbon to remove tastes and odours, and chlorination to kill the microbes from the sewage [1]

However, these stages do not remove dissolved nitrates from the farm run-off, so the treated water could still contain them and additional treatment would be required [1]

A reasoned overall judgement, for example: the river works would produce far more water for the energy used than a solar still, but its water quality depends on controlling the pollution upstream [1]

⚠ If you missed marks here: An evaluation must weigh advantages against limitations and end with a judgement. The specific chemical insight worth having is that the standard four stages remove suspended solids, tastes and microbes but leave dissolved nitrates untouched, whereas distillation removes everything. Listing the stages without discussing what they cannot do scores only the first mark.

1 mark -- names treatment stages and what each removes 1 mark -- dissolved nitrates not removed by these stages 1 mark -- reasoned overall judgement comparing the two options

Self-Assessment

Tick marks earned, then click Calculate Grade.

Your Score

Out Of

Percentage