AQA GCSE Combined Science: Trilogy Chemistry Paper 1 (Higher), 2020: Question 6

✅ Correct answers

• Atoms have a positively charged nucleus.
• Mass is concentrated in the nucleus in the centre of atoms.

Each correct tick = 1 mark. You needed two conclusions that match the deflection of some alpha particles and the fact that most passed straight through.

💡 Key knowledge

• Alpha particles are positively charged.
• They are repelled by a small, dense, positively charged nucleus.
• Most particles pass through because atoms are mostly empty space.
• Large deflections show that most of the mass and positive charge are concentrated in the nucleus.

🧠 Exam technique

• Pick statements that explain both the straight-through particles and the deflected ones.
• Use the experiment results, not just memorised atom facts.
• Look for wording about the nucleus, positive charge and mass concentrated.

❌ Common errors

• “Atoms are tiny spheres that cannot be divided” is the old model and is not supported by the results.
• “Atoms are balls of positive charge with embedded electrons” is not Rutherford’s model.
• “Neutrons exist within the nucleus” may be true, but it is not a conclusion from this experiment.

📐 Calculations: step by step

1. Use the thickness of the foil: 4.00 × 10⁻⁷ m
2. Divide by the number of atoms thick: 2400
3. 4.00 × 10⁻⁷ ÷ 2400 = 1.6666... × 10⁻¹⁰
4. Round to 3 significant figures: 1.67 × 10⁻¹⁰ m

Answer: diameter of one gold atom = 1.67 × 10⁻¹⁰ m

✅ Correct answer

1.67 × 10⁻¹⁰ m

This was worth 3 marks: method, correct calculation, and correct rounding/answer.

💡 Key knowledge

• Thickness = total thickness of foil.
• If the foil is 2400 atoms thick, one atom’s diameter is total thickness ÷ 2400.
• Always keep units in metres for the final answer.

❌ Common errors

• Forgetting to divide by 2400.
• Writing the answer as 1.67 × 10⁻¹⁰ cm or another incorrect unit.
• Giving too many/few significant figures.
• Using 4.00 × 10⁻⁷ as if it were the diameter of one atom, not the total thickness.

📐 Calculations: step by step

1. Start with the balanced equation: 2 Au + 3 Cl₂ → 2 AuCl₃
2. Find moles of gold: moles = mass ÷ Ar
3. 0.175 ÷ 197 = 0.000888 mol (or 8.88 × 10⁻⁴ mol )
4. Use the ratio 2 Au : 3 Cl₂
5. moles of Cl₂ = 0.000888 × 3 ÷ 2 = 0.00133 mol
6. Find mass of chlorine: mass = moles × Mr
7. 0.00133 × 71 = 0.0946 g
8. Convert to mg: 0.0946 g = 94.6 mg

Answer: 94.6 mg

✅ Correct answer

94.6 mg

This was worth 5 marks. The marks are for: finding moles of gold, using the ratio correctly, finding mass in grams, and converting to mg.

💡 Key knowledge

• Relative atomic mass: Au = 197, Cl = 35.5, so Cl₂ = 71.
• Balanced equation ratios matter: 2 mol Au react with 3 mol Cl₂.
• Use mass = moles × Mr .
• 1 g = 1000 mg.

🧠 Exam technique

• Always start with the balanced equation.
• Convert the given mass to moles first.
• Use the coefficient ratio carefully: × 3/2 for chlorine.
• Show units at every stage to avoid losing method marks.

❌ Common errors

• Using Cl instead of Cl₂ for the molar mass.
• Using the ratio the wrong way round, e.g. × 2/3 .
• Using 35.5 instead of 71 for chlorine gas.
• Forgetting to convert g to mg at the end.
• Not giving a final answer with the correct unit.

✅ What top answers did

• Linked scattering to a positive nucleus and concentrated mass.
• Used division to find one atom’s diameter.
• Used the balanced equation and correct mole ratio.
• Converted final mass into mg and gave the right number of significant figures.

❌ Where students lost marks

• Confusing Rutherford’s model with the plum pudding model.
• Missing the ratio step in the calculation.
• Using atomic mass of chlorine atoms, not chlorine molecules.
• Not converting to mg.