AQA GCSE Chemistry Chemistry Paper 2 (Foundation), June 2022: Question 4

✅ Correct Answer

Add two single covalent bonds:

• One single line connecting the right-hand carbon to the oxygen: C—O
• One single line connecting the oxygen to the hydrogen: O—H

❌ Common Errors

• Leaving the hydrogen joined directly to oxygen without a bond line (writing —OH instead of —O—H ). A displayed formula must show every single bond.
• Drawing double bonds to oxygen. Carbon forms 4 bonds, Oxygen forms 2, and Hydrogen forms 1.

✅ Correct Answer

Tick the second box:

☑ In hand gel to kill microbes

💡 Key Knowledge

• Ethanol destroys pathogens by denaturing their proteins and dissolving lipid membranes.
• Distractors: Aluminium protects itself naturally with an oxide layer (Al₂O₃), and testing for hydrogen gas uses a burning splint (squeaky pop test).

✅ Ideal 4-Mark Response

1. Draw a baseline in pencil across the chromatography paper (roughly 1–2 cm from the bottom).
2. Place a small spot of the ink onto the pencil start line.
3. Pour a suitable solvent (e.g., water or ethanol) into the beaker.
4. Suspend the paper in the beaker ensuring the solvent level is below the pencil line.
5. Place a lid on the beaker to prevent evaporation of the solvent.
6. Allow the solvent to rise up the paper until near the top, remove it, allow to dry, and observe the 2 distinct separated spots.

🧠 Exam Technique & Examiner Insights

• Level 3 (3–4 marks): A logical, sequenced method that would successfully produce the chromatogram shown in Figure 4.
• Crucial detail: State that the solvent must be below the baseline. If the solvent covers the line, the ink dissolves into the beaker wash rather than traveling up the paper!
• Why pencil? Ink contains dyes that would dissolve and run; pencil graphite is insoluble.

✅ Correct Answer

Yeast

💡 Key Knowledge

Fermentation reaction:

Glucose → Ethanol + Carbon Dioxide

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

Yeast provides the natural enzymes necessary to convert sugars to ethanol under warm (around 30–35 °C) and anaerobic (no oxygen) conditions.

💡 Alternative Order of Working

Convert units first, then find the percentage:

• 4.4 kg × 1000 = 4400 g (1 mark)
• 4400 g × 0.05 = 220 g (2 marks)

❌ Common Calculation Traps

• Forgetting unit conversion: Stopping at 0.22 and writing it on the line loses the conversion mark. Look carefully at the printed unit ( g ).
• Wrong conversion factor: Multiplying or dividing by 100 instead of 1000.
• Picking the wrong column: Using 95% (petrol) instead of 5% (ethanol).

✅ 3-Point Model Answer

1. Comparison: E10 contains a higher percentage of ethanol than E5 (10% vs 5%). (1 mark)
2. Crops required: Therefore, more sugar cane / sugar beet / plants must be grown to produce the fuel. (1 mark)
3. Mechanism: More carbon dioxide (CO₂) is absorbed from the atmosphere by these plants during photosynthesis as they grow. (1 mark)

🧠 Examiner Insights

• Link your ideas: Don't just say "plants take in CO₂". You must state that more plants/crops are grown because E10 needs more ethanol.
• Always name the biological process involved: photosynthesis ( 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ ).

✅ Acceptable Answers (Any 1)

• E10 releases less energy per unit mass / volume / per kg.
• The car will travel a shorter distance on a full tank (reduced fuel economy / lower miles per gallon).
• The driver needs to refuel more frequently.

💡 Using Table 7 Data

• Ethanol delivers 30.0 MJ/kg, whereas petrol delivers 46.4 MJ/kg.
• Because E10 contains 10% ethanol (twice as much low-energy fuel as E5's 5%), its total energy per kilogram is lower.