Edexcel A-Level Chemistry AS Paper 1, June 2016: Question 5

✅ Correct Answer

Mean titre = 19.95 cm³

Calculation: (19.90 + 20.00) ÷ 2 = 19.95 cm³

🧠 Exam Technique

• Always select concordant titres (titres that are within 0.10 cm³ of each other). Titration 1 (20.50 cm³) is a rough titre and must be discarded.
• Average only the concordant values (Titration 2 and Titration 3).

❌ Common Errors

• Including the rough first titre in the mean calculation, which distorts the final average value and loses the mark.
• Failing to show working or rounding incorrectly before subsequent calculations.

✅ Correct Answer

Concentration = 100.5 g dm⁻³ (or 100.5 g dm⁻³ with evaluation stating it is suitable because the extra 0.5 g dm⁻³ is insignificant / within experimental error).

📐 Step-by-Step Calculation

1. Moles of NaOH:
   n = (19.95 × 0.0800) / 1000 = 1.596 × 10⁻³ mol
2. Moles of diluted HNO₃ in 25.0 cm³:
   1:1 molar ratio means n(HNO₃) = 1.596 × 10⁻³ mol
3. Concentration of diluted HNO₃ (in mol dm⁻³):
   c = (1.596 × 10⁻³ / 25.0) × 1000 = 6.384 × 10⁻² mol dm⁻³
4. Concentration of undiluted HNO₃ (accounting for dilution factor):
   Diluted from 10.00 cm³ to 250 cm³ (factor of 25).
   6.384 × 10⁻² × (250 / 10.00) = 1.596 mol dm⁻³
5. Convert to g dm⁻³ (Molar mass of HNO₃ = 63.0 g mol⁻³):
   1.596 × 63.0 = 100.548 g dm⁻³ = 100.5 g dm⁻³ (to 1 decimal place)

💡 Key Knowledge & Guidance

• Allowances: Titre volumes between 19.9 and 20.5 cm³ are accepted if based on student's own calculated mean.
• Consequential Error (TE): Errors carried forward from part (a) or rounding steps are fully credited provided method is sound.
• The final mark requires both the correct numerical answer to 1 d.p. and a valid evaluative statement regarding suitability.

💡 Key Structural Requirements

• Hydrogen & Left Oxygen: Single covalent bond between H and O (H has 2 electrons, O has octet).
• Nitrogen & Second Oxygen: Double covalent bond between N and one of the terminal oxygens (N=O).
• Third Oxygen: Single covalent bond linking N to the remaining -OH group or dative/covalent single bond with lone pairs.
• Strictly use the designated symbols: (•) for oxygen electrons, (x) for nitrogen electrons, and (*) for the hydrogen electron.

❌ Common Misconceptions

• Confusing electron symbols or failing to show all outer shell lone pairs on oxygen atoms.
• Drawing incorrect valencies (Nitrogen must have an octet share around it; remember N forms 3 bonds typically in uncharged structures, but here has a dative/coordination or double bond arrangement totalling 5 bonds/shared pairs in hypervalent/resonance depictions accepted by Edexcel). Follow the canonical mark scheme: N double-bonded to one O, single-bonded to OH, and single-bonded to the third O.

✅ Correct Answer

Atom Economy = 80.8%

📐 Calculation & Formula

Atom Economy = (Mass of desired product / Total mass of reactants) × 100

• Desired product (4 HNO₃): 4 × (1 + 14 + (16×3)) = 4 × 63 = 252 g
• Total reactants (3 N₂O₄ + 2 H₂O):
  Mass of 3(N₂O₄) = 3 × ((14×2) + (16×4)) = 3 × 92 = 276 g
  Mass of 2(H₂O) = 2 × ((1×2) + 16) = 2 × 18 = 36 g
  Total Reactant Mass = 276 + 36 = 312 g  *(Wait, let's look at MS: 252 / (252 + 60) where 60 is mass of 2NO byproduct).*
  Alternatively using Products: 252 (HNO₃) + 60 (2NO byproduct) = 312 g total mass.
• Calculation: (252 / 312) × 100 = 80.769% = 80.8%

❌ Examiner Warning

• Do not award 80.7% if improperly truncated without proper rounding. The mark scheme specifically accepts 80.8(%) and instructs: "Ignore sf except one. Do not award 80.7%".