AQA GCSE Combined Science: Trilogy Chemistry Paper 1 (Higher), 2023: Question 2

✅ Correct answer (what gets the mark)

Neutralisation

💡 Key knowledge

• Acid + alkali → salt + water is neutralisation.
• Here: sulfuric acid (acid) + lithium hydroxide (alkali) → lithium sulfate (salt) + water.

🧠 Exam technique

• If you see acid + hydroxide and water is made, write neutralisation.
• Keep it simple: one-word answer is enough for 1 mark.

❌ Common errors (examiner insight)

• Writing only “exothermic”. The mark scheme says this is allowed, but the expected answer is neutralisation. Don’t risk it—use the reaction type.
• Calling it “acid reaction” or “salt formation” (too vague).

📐 Calculations (step-by-step)

1. Count atoms in H₂SO₄: H = 2, S = 1, O = 4.
2. Multiply each by Aᵣ:
   • H: 2 × 1 = 2
   • S: 1 × 32 = 32
   • O: 4 × 16 = 64
3. Add them: 2 + 32 + 64 = 98

✅ Correct answer + marks

Mᵣ(H₂SO₄) = 98

❌ Common errors (why marks are lost)

• Forgetting the 2 in H₂ (using 1 × 1 instead of 2 × 1).
• Forgetting the 4 in O₄ (using 1 × 16 instead of 4 × 16).
• Writing units (Mᵣ has no units).

🧠 Exam technique

• Always write a “sum line” like (2 × 1) + 32 + (4 × 16) to secure the method mark.
• Underline the subscripts (2 and 4) on the formula before you start.

📐 Calculations (step-by-step)

1. Find the total mass of oxygen in Li₂SO₄:
   O₄ means 4 oxygen atoms → 4 × 16 = 64
2. Use the percentage by mass formula:
   percentage of O = (mass of O in formula ÷ Mᵣ of compound) × 100
3. Substitute values:
   (64 ÷ 110) × 100 = 58.18...
4. Round to 2 significant figures:
   58%

✅ Correct answer + marks

Percentage by mass of oxygen = 58% (2 s.f.)

Examiner note: the scheme allows correct method even if a student’s oxygen mass is wrong, as long as they then use their value consistently (error carried forward).

💡 Key knowledge

• % by mass = (mass of element in 1 mole of compound ÷ Mᵣ of compound) × 100
• Subscripts matter: O₄ means 4 oxygen atoms.
• 2 significant figures for 58.18 is 58 (not 58.2).

❌ Common errors (examiner insight)

• Using 16 instead of 4 × 16 (forgetting there are 4 oxygens).
• Dividing the wrong way round: 110/64 × 100 (gives >100%, which should be an instant red flag).
• Rounding mistakes: giving 58.18% (not 2 s.f.) or 58.2% (that’s 3 s.f.).

🧠 Exam technique

• Quick sense-check: oxygen is a big part of Li₂SO₄, so a value around 50–60% seems realistic.
• Write your calculator result, then clearly show the rounding step to 2 s.f. to protect the final mark.

📐 Calculations (method that matches the mark scheme)

1. Convert volume to dm³:
   25 cm³ ÷ 1000 = 0.025 dm³
2. Use: concentration = mass ÷ volume
   0.30 ÷ 0.025 = 12
3. Add the correct unit:
   12 g/dm³

✅ Correct answer

Concentration = 12 g/dm³

Examiner note: the mark scheme also accepts an alternative route via g/cm³ then converting to g/dm³.

💡 Key knowledge

• 1 dm³ = 1000 cm³
• So to convert cm³ → dm³, divide by 1000.
• Concentration in g/dm³ means “grams per dm³”.

🧠 Exam technique

• Always write the conversion line ( 25 ÷ 1000 )—that’s often a dedicated mark.
• Estimate: 25 cm³ is 1/40 of a dm³, so the concentration should be about 40 × 0.30 ≈ 12 g/dm³ (good check).

❌ Common errors (calculation traps)

• Converting the wrong way: multiplying by 1000 gives 25,000 dm³ (nonsense).
• Forgetting units in the final answer (you need g/dm³).
• Dividing by 25 instead of 0.025 (mixing cm³ and dm³).

🧠 To hit full marks, make sure you…

• Use the exact keyword neutralisation for part (a).
• Show an explicit Mᵣ sum for part (b), not just the final number.
• For % by mass, write (mass of element / Mᵣ) × 100 and then round to 2 s.f.
• Convert cm³ → dm³ before using g/dm³.

💡 Mini recall

• Neutralisation: acid + alkali → salt + water
• Mᵣ: add up atomic masses using subscripts
• 1 dm³ = 1000 cm³