Water Worked ExamplesIGCSE Chemistry 0620

Pure water boils at exactly $100\,^{\circ}\text{C}$ (and melts at exactly $0\,^{\circ}\text{C}$). Dissolved substances such as salts make water an impure substance, so its boiling point is higher than $100\,^{\circ}\text{C}$. This rules out A and B.

Anhydrous cobalt(II) chloride is blue. Water from any source — pure or not — turns it pink, because the test shows only that water is present. So the colour change is blue to pink, which rules out D (the reverse change, seen when pink hydrated cobalt(II) chloride is heated).

Answer: C.

(a) Dissolved oxygen is needed by aquatic life (fish and other organisms breathe it). Calcium compounds are metal compounds that provide essential minerals for life.

(b)(i) Sewage contains harmful microbes which cause disease. (ii) Some plastics harm aquatic life (for example animals swallow them or become trapped). (iii) Some metal compounds, such as lead compounds, are toxic.

Nitrates and phosphates are also harmful: they lead to deoxygenation of the water and damage to aquatic life. You are not asked to describe the steps of that process.

(c) Phosphates also come from detergents (in waste water from homes and laundries).

(a) Tank A is sedimentation: large insoluble particles (sand, mud) settle to the bottom under gravity and are separated from the water.

(b) Stage B is filtration through beds of sand. It removes the smaller insoluble solids still suspended in the water.

(c) Carbon removes substances that cause tastes and odours.

(d) Chlorination kills microbes (bacteria), so the water does not spread disease.

(e) The solids are removed first so the chlorine acts on clear water; adding it last means the microbes are killed just before the water enters the pipes to homes, so the supply stays safe to drink.

The syllabus gives the two hydrated salts you must know: $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$ and $\text{CoCl}_2\cdot 6\text{H}_2\text{O}$. Cobalt(II) chloride has six molecules of water of crystallisation, which rules out A (five belongs to copper(II) sulfate).

Anhydrous cobalt(II) chloride is blue and turns pink when water is added. C gives the reverse change (what happens on heating the hydrated salt). D is the colour change for anhydrous copper(II) sulfate, not cobalt(II) chloride.

Answer: B.

(a) Heating drives off the water of crystallisation, leaving white anhydrous copper(II) sulfate.

(b) Five molecules of water are lost per formula unit. Check: Cu 1 = 1, S 1 = 1, O 9 = 4 + 5, H 10 = 10.

The water leaves as steam and condenses to liquid in the cooled test tube.

(c) Add a few drops of Y to anhydrous copper(II) sulfate: it turns from white to blue. (Or: anhydrous cobalt(II) chloride turns from blue to pink.)

(d) Measure the boiling point: pure water boils at exactly $100\,^{\circ}\text{C}$ (or check that it freezes/melts at exactly $0\,^{\circ}\text{C}$).

(e) The reaction reverses: the white solid turns blue and the mixture gets hot (the hydration is exothermic, as the dehydration needed heat).

(f) Hydrated. (A substance containing no water is anhydrous; the water molecules present in hydrated crystals are the water of crystallisation.)

(a) Add the anhydrous part and six waters.

(b) Percentage of water:

(c) Moles of hydrated salt; each mole gives one mole of anhydrous $\text{CoCl}_2$ ($M_r = 130$).

Check: mass of water lost $= 11.9 - 6.50 = 5.40$ g, and $\dfrac{5.40}{11.9} \times 100 = 45.4\%$, agreeing with (b).

(d) Hydrated cobalt(II) chloride is pink; the anhydrous salt is blue. So the solid turns from pink to blue.

(a) The mass fell between the first and second heatings, so water was still being lost. The student heats until two readings agree (constant mass) — this shows all the water of crystallisation has been removed.

(b) Use the final constant mass.

(c) Convert both masses to moles. $M_r(\text{MgSO}_4) = 24 + 32 + 4(16) = 120$; $M_r(\text{H}_2\text{O}) = 18$.

Divide by the smaller number of moles to get the ratio.

So the formula is $\text{MgSO}_4\cdot 7\text{H}_2\text{O}$.

(d) Lifting the lid lets the steam escape; keeping it on the crucible stops solid spitting out, which would lose mass that is not water.

(a) Sample P. It boils at a fixed temperature of exactly $100\,^{\circ}\text{C}$ (the line is flat at 100). Sample Q boils above $100\,^{\circ}\text{C}$ and its temperature keeps rising — it boils over a range, which shows it is impure.

(b) Anhydrous copper(II) sulfate turns blue with any water, pure or impure. It shows that water is present, not that it is pure.

(c) Acidify the sample with dilute nitric acid, then add aqueous silver nitrate. A white precipitate shows chloride ions.

(d) The precipitate is silver chloride. Charges balance: $+1 - 1 = 0$.

(a) The method is (simple) distillation. X is a (Liebig) condenser.

(b) Water entering at the bottom fills the jacket completely, so the condenser is as cold as possible along its whole length and all the steam condenses.

(c) $100\,^{\circ}\text{C}$ — the thermometer bulb is at the side arm, so it reads the boiling point of the vapour, which is pure water. (The salts stay in the flask.)

(d) Distilled water contains fewer chemical impurities than tap water. Tap water contains dissolved substances (for example chloride ions from chlorination, or metal compounds) that could react or give false test results.

(e) Boiling large volumes of water needs a great deal of energy (heat), so it is expensive; river water can be treated by sedimentation, filtration and chlorination instead.

(a) Convert mg to g ($\div 1000$), then g to mol ($M_r(\text{O}_2) = 32$).

(b) Moles in $5.0 \text{ dm}^3$, then volume at r.t.p.

(c) Percentage decrease = decrease ÷ original × 100.

(d) Nitrates and phosphates from fertilisers washed off the fields. They lead to deoxygenation of the water, and aquatic life (fish) that needs dissolved oxygen is damaged or dies.

(a) The damp red litmus paper turns blue. The gas is ammonia, $\text{NH}_3$ — an alkaline gas. (Nitrate ions are reduced by aluminium in the alkaline solution.)

(b) Warm a fresh sample with aqueous sodium hydroxide without the aluminium foil. If the damp red litmus stays red, no ammonium ions are present, so the ammonia in the full test must come from nitrate ions.

(c) $M_r(\text{NO}_3^{\,-}) = 14 + 3(16) = 62$. Convert mol/dm³ to g/dm³, then to mg/dm³.

$74.4 \text{ mg/dm}^3$ is greater than $50 \text{ mg/dm}^3$, so the water is not safe to drink without treatment.

(a) Mass = concentration × volume, then convert units ($1 \text{ kg} = 10^6 \text{ mg}$).

(b) $M_r(\text{Cl}_2) = 2 \times 35.5 = 71$.

(c) Use $24 \text{ dm}^3$ per mole (keep the unrounded moles).

(d) Chlorine kills microbes in the water. Tastes and odours are removed by treatment with carbon.