Edexcel A-Level Chemistry Paper 1, November 2020: Question 4

8 marks · Hard difficulty · Extended Writing

Assess the solubility of Group 2 sulfates, calculate the enthalpy change of solution for magnesium chloride using a Born-Haber / enthalpy cycle, and explain the solubility of 2-methylpentane and potassium bromide in water and hexane based on intermolecular forces.

Practise this question

Question

Question 4 asks about dissolving compounds in three parts. Part (a) is a multiple-choice question on the solubility of Group 2 sulfates. Part (b) provides an enthalpy cycle for magnesium chloride and asks for the standard enthalpy change of solution from given lattice energy and hydration enthalpy values. Part (c) gives a table showing the solubility of 2-methylpentane and potassium bromide in water and hexane, and asks for a 6-mark extended response explaining the findings based on intermolecular interactions.
Question text

4 This question is about dissolving different compounds.

(a) Which of these compounds is the most soluble in water?

(1)

A barium sulfate

B calcium sulfate

C magnesium sulfate

D strontium sulfate

(b) What is the value, in kJ mol−1, for the standard enthalpy change of solution of

magnesium chloride?

Mg2+(g) + 2Cl−(g)

MgCl (s) Mg2+(aq) + 2Cl−(aq)

Lattice energy MgCl (s) = −2526 kJ mol−1

Hydration enthalpy of Cl−(g) = − 381 kJ mol−1

Hydration enthalpy of Mg2+(g) = −1921 kJ mol−1

(1)

A +157

B −157

C +224

D −224

*(c)The solubility of two compounds in different solvents was investigated.

A summary of the findings is shown.

Compound Soluble in water Soluble in hexane

2-methylpentane X

potassium bromide X

Explain the findings of the investigation by considering the interactions between

the compounds and each of the solvents.

(6)

… *P62668A01124*

… *P62668A01224*

(Total for Question 4 = 8 marks)

Mark scheme

Show the mark scheme The mark scheme provides the correct multiple-choice answers for parts (a) and (b), and a detailed breakdown for part (c) including indicative content points covering hydrogen bonding, London forces, ionic hydration, and lattice enthalpies alongside structure and communication marks.

How to answer it

Dissolving, Enthalpy Cycles, and Intermolecular Forces

What this question tests

This question assesses your understanding of Group 2 trends in solubility, application of Hess's Law using Born-Haber / enthalpy cycles involving lattice energy and hydration enthalpies, and explaining solubility in terms of intermolecular forces and ionic hydration ("like dissolves like").

Question 4(a) - Solubility Trends

Group 2 Sulfates Solubility

✅ Correct Answer

C (magnesium sulfate)

1 Mark

💡 Key Knowledge

  • The solubility of Group 2 sulfates decreases down the group (from magnesium down to barium).
  • Magnesium sulfate (MgSO₄) is the most soluble among the options provided (barium, calcium, magnesium, and strontium sulfates).
Question 4(b) - Enthalpy of Solution Calculation

Standard Enthalpy Change of Solution for Magnesium Chloride

✅ Correct Answer

B (-157 kJ mol⁻¹)

1 Mark

📐 Step-by-Step Calculation

  1. Write the Hess's Law route: ΔH_sol = ΣΔH_hydration - ΔH_lattice
  2. Identify values from the diagram:
    Lattice energy of MgCl₂(s) = -2526 kJ mol⁻¹
    Hydration enthalpy of Mg²⁺(aq) = -1921 kJ mol⁻¹
    Hydration enthalpy of Cl⁻(aq) = -381 kJ mol⁻¹
  3. Account for stoichiometry: There are 2 moles of chloride ions, so multiply the Cl⁻ hydration enthalpy by 2:
    2 × (-381) = -762 kJ mol⁻¹
  4. Calculate total hydration enthalpy: (-1921) + (-762) = -2683 kJ mol⁻¹
  5. Apply the cycle equation: ΔH_sol = (-2683) - (-2526) = -157 kJ mol⁻¹

❌ Common Calculation Traps

  • Forgetting stoichiometry: Forgetting to multiply the Cl⁻ hydration enthalpy by 2 leads to distractor C (+224).
  • Sign errors / wrong direction: Using the cycle backwards results in positive values like +157 (Distractor A) or +224.
Question 4(c) - Intermolecular Forces & Solubility

Explaining Solubility Findings (6 Marks)

💡 Key Knowledge (Indicative Content)

  • 2-methylpentane (non-polar): Cannot form hydrogen bonds with water because it lacks sufficiently electronegative atoms attached to hydrogen.
  • 2-methylpentane in hexane: Soluble because both are non-polar molecules sharing similar London (dispersion) forces of comparable strength/size. Resulting forces in the mixture match those in pure liquids.
  • Potassium bromide (ionic): Soluble in water because ion-dipole bonds form and ions are effectively hydrated.
  • Enthalpy compensation: The energy released during hydration of K⁺ and Br⁻ ions is sufficient to compensate for the energy needed to break the ionic lattice.
  • Potassium bromide in hexane: Insoluble because any London forces forming between the ionic lattice and non-polar hexane are vastly too weak to overcome strong electrostatic forces holding the ionic lattice together.

🧠 Exam Technique & Mark Structure (6 Marks Total)

This is a Levels of Response question evaluated across two criteria:

  • Indicative Content (IC - up to 4 marks): Awarded based on how many correct chemical points (bullet points above) you successfully make.
  • Lines of Reasoning (RoR - up to 2 marks): Awarded for a coherent, logical structure linking the properties of both compounds in both solvents with clear terminology (e.g. explicitly mentioning intermolecular forces, polarity, and hydration).
Top-level responses explicitly contrast polar vs non-polar interactions and avoid vague phrases like "like dissolves like" without explaining *why* (e.g. referencing hydrogen bonding or London forces).

Topics

Physical Chemistry · Inorganic Chemistry · Organic Chemistry · Topic 13: Energetics II · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 2: Bonding and Structure

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.