Edexcel A-Level Chemistry Paper 2, June 2022: Question 5

5 marks · Medium difficulty · Calculations

Deduce the density of cooking oil based on layer formation and calculate the difference in number of molecules between given volumes of water and ice.

Practise this question

Question

The question presents two parts about the densities of ice and water. Part (a) describes mixing water and cooking oil with an ice cube, asking for a possible value for the density of cooking oil with justification (2 marks). Part (b) asks to calculate how many more molecules are in 5.00 cm3 of water compared to 5.00 cm3 of ice (3 marks).
Question text

5 Ice has a density of 0.92 g cm–3 and water has a density of 1.00 g cm–3.

(a) About 200 cm3 of water and 200 cm3 of cooking oil were placed in a large beaker

and two layers formed. The cooking oil formed the upper layer.

An ice cube made from water with a water-soluble blue food dye was added.

Initially the ice cube floated on top of the cooking oil but on melting the

blue-coloured water sank into the bottom layer of water.

Give a possible value for the density of the cooking oil. Justify your answer.

(2)

(b) Calculate how many more molecules there are in 5.00 cm3 of water compared to

5.00 cm3 of ice.

(3)

(Total for Question 5 = 5 marks)

Mark scheme

Show the mark scheme The mark scheme for part 5(a) awards 1 mark for a density value between 0.92 and 1.00 g cm-3 and 1 mark for justifying using layer positions and ice floating. For part 5(b), 3 marks are awarded for using both densities to find masses or moles, subtracting them or calculating molecules separately before subtraction, and multiplying by the Avogadro constant to yield 1.34 x 10^22.

Question

Answer Additional Guidance Mark

Number

5(a) An answer which makes reference to the following points: (2)

• density between 0.92 and 1.00 (g cm−3) (1) Accept any value or range between

0.92 − 1.00

Ignore units even if incorrect

• because water is the bottom layer so more dense Accept reverse arguments

and Reference to the layers is required

ice floats on oil so is less dense (1)

Question

Answer Additional Guidance Mark

Number

5(b) An answer that give evidence of the following: Multiple correct methods are possible which process (3)

the data in different sequences.

The correct final answer is 1.34 x 1022 / 1.338 x 1022

which can be awarded (3) regardless of working

If this answer is not given then look for evidence of

each of the given mathematical processes and give

one mark for each

• use of both densities to get two masses The use of both densities must be carried out first

and Note that the use of 5 for the mass of water implies

division by 18 to give moles (1) the use of a density of 1.00 g cm−3

• subtraction to give either mass or moles Depending on the method used this can be done at

or number of molecules (1) the beginning, the middle or at the end of the

calculation but must be of (water – ice)

• multiplication by Avogadro constant to This must be evidenced after moles have been

give number of molecules (1) calculated

Allow TE throughout

Ignore SF except 1SF for the final answer

Allow use of 6 x 1023 which gives 1.33 x 1022 for (3)

Correct answer without working scores (3)

Do not allow a number of molecules <1

Marking points Example of calculation vs1

Subtraction (1) m(water) = (5 x 1.00) − (5 x 0.92) = 0.40 (g)

Use of both densities and n(H2O) = (0.40 ÷ 18)

division by 18 to give moles (1) = 0.022222 / 2.2222 x 10−2 (mol)

Multiplication by Avogadro N = (2.2222 x 10−2 x 6.02 x 1023)

constant (1) =1.34 x 1022 / 1.338 x 1022

or Example of calculation vs2

Multiplication by Avogadro N(water molecules) = ((5x 1) ÷ 18) x 6.02 x 1023

constant (1) = 1.667 x 1023

Use of both densities and N(ice molecules) = ((5 x 0.92) ÷ 18) x 6.02 x 1023

division by 18 to give moles (1) = 1.533 x 1023

Subtraction (1) N(Extra) = 1.667 x 1023 − 1.533 x 1023 = 1.34 x 1022

or Example of calculation vs3

Use of both densities and n(water) = ((1.00 x 5.00) ÷ 18) = 0.27778 (mol)

division by 18 to give moles (1) n(ice) = ((0.92 x 5.00) ÷ 18) = 0.25556 (mol)

Subtraction (1) Difference in mol = (0.27778 − 0.25556)= 0.022222(mol)

Multiplication by Avogadro Extra molecules = 0.022222 x 6.02 x 1023 = 1.34 x 1022

constant (1)

(Total Question 5 = 5 marks)

How to answer it

Properties of Matter: Density & Moles Study Guide

What this question tests

This question assesses your understanding of physical properties (specifically density and floating/sinking behaviour in immiscible liquids), molar calculations (mass = volume × density, moles = mass ÷ molar mass), and the application of the Avogadro constant to find the actual number of particles.

Question 5(a)

Density and Immiscible Layers

Task: Give a possible value for the density of the cooking oil and justify your answer.

✅ Correct Answer

Density value: Any value or range between 0.92 g cm⁻³ and 1.00 g cm⁻³ (exclusive or inclusive, e.g., 0.95 g cm⁻³ ).

Justification: Water is the bottom layer so it is more dense ( 1.00 g cm⁻³ ), and ice floats on oil so oil is less dense than ice ( 0.92 g cm⁻³ ).

💡 Key Knowledge

  • Substances with higher densities sink below substances with lower densities.
  • Ice floats on the oil initially, meaning ice density < oil density.
  • Water is the bottom layer beneath the oil, meaning oil density < water density.

❌ Common Errors

  • Giving a single number outside the 0.92 – 1.00 range.
  • Failing to mention *both* reference points (ice and water) in the justification.
  • Forgetting to state that layer positioning directly corresponds to relative density values.
Mark breakdown (2 marks): 1 mark for a density value between 0.92 and 1.00 g cm⁻³. 1 mark for the justification referencing the positions of water (bottom/more dense) and ice (floating on oil/less dense).
Question 5(b)

Calculating Molecular Difference

Task: Calculate how many more molecules there are in 5.00 cm³ of water compared to 5.00 cm³ of ice.

✅ Correct Answer

Final Answer: 1.34 × 10²² (or 1.338 × 10²² ) molecules.

Note: The mark scheme awards full marks (3/3) for the correct final answer regardless of the sequence of working steps used!

🧠 Exam Technique

  • Choose a calculation flow that suits you: you can find masses first, mole differences second, or find total molecules of each state separately and subtract at the end.
  • Keep full calculator values throughout intermediate steps to avoid rounding errors. Round your final answer to 2 or 3 significant figures.

📐 Step-by-Step Calculation (Method 1)

  1. Find the mass difference:
    Mass of water = 5.00 × 1.00 = 5.00 g
    Mass of ice = 5.00 × 0.92 = 4.60 g
    Mass difference = 5.00 - 4.60 = 0.40 g
  2. Convert mass difference to moles (H₂O = 18.0):
    Moles = 0.40 ÷ 18.0 = 0.02222 mol
  3. Multiply by the Avogadro constant:
    Molecules = 0.02222 × 6.02 × 10²³ = 1.34 × 10²²

❌ Common Calculation Traps

  • Using the density of water ( 1.00 g cm⁻³ ) for ice or vice-versa.
  • Forgetting that H₂O has a molar mass of 18.0 g mol⁻¹ (using 16.0 by forgetting hydrogens).
  • Giving an unrounded 1-significant-figure answer (e.g., 1 × 10²² is explicitly disallowed).
Mark breakdown (3 marks): 1 mark for using both densities to determine masses/moles. 1 mark for subtraction (water minus ice). 1 mark for multiplication by the Avogadro constant. (Allow ECF throughout).

Topics

Physical Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 2: Bonding and Structure

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.