Cambridge International AS & A Level Biology 9700: Water – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 2.4 Water (learning outcome 1) of Topic 2, Biological molecules, in the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It is AS Level content, examined on Papers 1 and 2, and Paper 4 can draw on it. Calculators are allowed in 9700 examinations. Question 1 is in multiple-choice style; the rest are structured.

Learn it first: Water study guide · Recall: Water revision notes · Whole-topic questions: Biological molecules practice questions · Course hub: Cambridge A Level Biology · Checklist: 9700 checklist

Questions

1. Which statement explains why water has a high latent heat of vaporisation? [1]

• A Covalent bonds inside water molecules must be broken before water can evaporate.
• B Hydrogen bonds between water molecules must be broken before water can evaporate.
• C Water molecules are non-polar, so they leave the liquid surface slowly.
• D Water has a high specific heat capacity, so it stores heat before evaporating.

2.

(a) Explain why a water molecule is polar. [2] (b) Describe how a hydrogen bond forms between two water molecules. [2]

3. Draw a labelled diagram of two water molecules joined by one hydrogen bond. Show the charges on the atoms. [3]

4. Root hair cells absorb potassium ions (K⁺) and chloride ions (Cl⁻) from the soil solution.

(a) Describe how water molecules are arranged around a K⁺ ion and around a Cl⁻ ion in solution. [2] (b) State why this property of water is important for the movement of mineral ions to the leaves. [1]

5. A student stirs a spoonful of sucrose into one beaker of water and a few drops of olive oil (mostly triglycerides) into another. The sucrose dissolves. The oil forms a separate layer on the surface. Explain these observations. [3]

6. An aquarium holds 40 kg of water. Over one afternoon the water absorbs 504 kJ of energy from a heater and sunlight, with no heat lost. The body temperature of the fish in the tank follows the water temperature.

(a) Calculate the rise in temperature of the water. Take the specific heat capacity of water as 4.2 J g⁻¹ °C⁻¹. Show your working. [2] (b) Suggest why the size of this temperature rise is important for the fish. [2]

7. During a long cycle ride, a cyclist must lose 1.8 MJ of heat by evaporation of sweat. Take the latent heat of vaporisation of water as 2.4 kJ g⁻¹.

(a) Calculate the mass of sweat that must evaporate. [2] (b) Only 75% of the sweat produced evaporates; the rest drips off. Calculate the total mass of sweat the cyclist must produce. [2] (c) Explain why sweat that drips off the skin does little to cool the cyclist. [1]

8. In an investigation, two plants of the same species were kept in bright light at the same air temperature. Plant P was well watered and its stomata were open. Plant Q was in dry soil and its stomata were closed. The leaf temperature of P was 29 °C and that of Q was 34 °C. Explain the difference in leaf temperature. [3]

9. Distinguish between the specific heat capacity of water and the latent heat of vaporisation of water. [2]

10. Water makes up most of the mass of a mammal’s cells and blood.

(a) Explain how the properties of water allow substances to be transported in blood plasma. [3] (b) Explain how hydrogen bonding between water molecules helps a mammal to keep a stable body temperature on a hot day. Refer to two different properties of water. [5]

11. This question compares the two thermal properties of water. Take the latent heat of vaporisation as 2.4 kJ g⁻¹ and the specific heat capacity as 4.2 J g⁻¹ °C⁻¹.

(a) Calculate the energy removed when 10 g of water evaporates. [1] (b) Calculate the fall in temperature this amount of energy would cause if it were removed from 2.0 kg of liquid water. Give your answer to 2 significant figures. [2] (c) A small desert mammal has very little water to spare. Use your answer to (a) to explain why evaporative cooling can still be useful to it. [2] (d) Suggest why panting cools a dog less well in very humid air. [2]

Answers

1. B [1] [1] Examiner insight: A is the classic distractor: the covalent O–H bonds are not broken when water evaporates, and choosing it shows the two bond types have been confused.

2. (a) Oxygen attracts the shared electrons in the O–H bonds more strongly than hydrogen, so oxygen is δ− and each hydrogen is δ+ [1]. The molecule is bent, so the charge is unevenly spread, with a δ− side and a δ+ side [1]. (b) The δ+ hydrogen of one water molecule [1] is attracted to the δ− oxygen of a neighbouring water molecule; this weak attraction is the hydrogen bond [1]. [4] Examiner insight: In (b), “the molecules attract each other” is not enough; the mark needs the specific atoms, δ+ H on one molecule and δ− O on the other.

3. Two molecules each drawn as one O joined to two H atoms by solid lines, in a bent shape [1]. δ− labelled on each O and δ+ on each H [1]. A dashed or dotted line from an H of one molecule to the O of the other, labelled hydrogen bond [1]. [3] Examiner insight: A dotted line drawn from O to O, or between two H atoms, loses the third mark even if it is labelled correctly.

4. (a) Around K⁺, water molecules turn their δ− oxygen towards the ion [1]. Around Cl⁻, water molecules turn their δ+ hydrogens towards the ion [1]. (b) The ions are dissolved, so they can be carried in the water moving up the xylem [1]. [3] Examiner insight: Both ions must be described for (a); a reversed answer (H towards K⁺) scores zero for that ion.

5. Sucrose has many –OH groups, so it is polar [1]. Its –OH groups form hydrogen bonds with water molecules, so it dissolves [1]. Triglycerides are non-polar and cannot form hydrogen bonds with water, so the oil does not dissolve and stays as a separate layer [1]. [3] Examiner insight: “Oil is hydrophobic” restates the observation; the third mark needs the reason, that it cannot form hydrogen bonds with water.

6. (a) temperature change = 504 000 ÷ (40 000 × 4.2) [1] = 3.0 °C [1]. (b) The fish’s body temperature changes only a little, so its enzymes stay near their optimum temperature [1] and are not denatured, so its rate of metabolism stays steady [1]. [4] Examiner insight: In (a), converting only one quantity (kg to g, or kJ to J) gives an answer out by a factor of 1 000 and loses the accuracy mark; the method mark can still be awarded.

7. (a) mass = 1 800 ÷ 2.4 [1] = 750 g [1]. (b) total = 750 ÷ 0.75 [1] = 1 000 g (1.0 kg) [1]. (c) Heat is only removed from the skin when water evaporates, by breaking hydrogen bonds; liquid sweat that drips off has not taken in the latent heat of vaporisation [1]. [5] Examiner insight: In (b), multiplying 750 by 0.75 instead of dividing is a common slip; the answer must be larger than (a), and a value smaller than 750 g scores no marks.

8. In P, water evaporates from the mesophyll cell walls and leaves through open stomata in transpiration [1]. Evaporation needs a lot of energy because hydrogen bonds between water molecules must break, as water has a high latent heat of vaporisation [1]. This energy is taken from the leaf, cooling it; in Q the stomata are closed, so there is little evaporation and little cooling [1]. [3] Examiner insight: An answer that only says “P loses more water” misses the point; marks come from linking evaporation, latent heat and the transfer of energy away from the leaf.

9. Specific heat capacity is the energy needed to raise the temperature of 1 g (or 1 kg) of liquid water by 1 °C [1]. Latent heat of vaporisation is the energy needed to change liquid water into water vapour, with no change in temperature [1]. [2] Examiner insight: Each definition must say what changes, a rise in temperature for one and a change of state for the other, or the two marks merge into one.

10. (a) Water molecules are polar [1]. They surround ions and form hydrogen bonds with polar molecules such as glucose and amino acids, so these substances dissolve [1]. Dissolved substances are carried in the plasma as the blood flows round the body [1]. (b) Water molecules are held together by many hydrogen bonds [1]. Much energy is used to break hydrogen bonds before the temperature rises, so water has a high specific heat capacity [1]. So the body’s temperature rises only slowly as heat is gained [1]. When sweat evaporates, hydrogen bonds must break, so water has a high latent heat of vaporisation [1]. The evaporating water takes a large amount of energy from the skin, cooling the body with little water lost [1]. [8] Examiner insight: In (b), credit for each property needs its own hydrogen-bond reason and its own effect on the mammal; naming the two properties without explanation earns little.

11. (a) 10 × 2.4 = 24 kJ [1]. (b) 24 000 ÷ (2 000 × 4.2) [1] = 2.9 °C [1]. (c) Evaporating a small mass of water removes a large amount of energy, as (a) shows [1], so the mammal can lose enough heat while losing little water [1]. (d) Humid air already holds a lot of water vapour, so less water evaporates from the mouth and airways [1]. Less evaporation means less energy is removed, so less cooling [1]. [7] Examiner insight: In (b), 2.857 must be rounded to 2.9 as asked; giving 2.86 or 3 loses the accuracy mark, though the method mark stands.

Where marks are usually lost

• Calling the O–H bond inside a water molecule a hydrogen bond, or picking that idea in a multiple-choice item.
• Describing a hydrogen bond without naming the δ+ hydrogen and δ− oxygen on different molecules.
• Reversing the orientation of water molecules around positive and negative ions.
• Writing “hydrophobic” as the explanation instead of “cannot form hydrogen bonds with water”.
• Mixing kg with g, or kJ with J, in thermal calculations.
• Dividing where a multiplication is needed (or the reverse) when a percentage of sweat evaporates.
• Linking transpiration only to water loss, not to energy removed from the leaf.
• Giving a property with no role, or a role with no hydrogen-bond reason.
• Saying high specific heat capacity keeps organisms “warm” rather than keeping temperature stable.

Next steps

• Water revision notes
• Water study guide
• Biological molecules practice questions
• Cambridge A Level Biology hub
• 9700 printable checklist
• Free 9700 AS diagnostic
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment – Topic 2, Biological molecules (section 2.4 Water).