📖 What's the chemistry?
Q1 — water of crystallisation (gravimetry)
CuSO₄·xH₂O(s) → CuSO₄(s) + xH₂O(g). Heating drives off the water of crystallisation, so
the mass falls; the moles of water lost divided by the moles of anhydrous CuSO₄ left gives x.
- Heat gently — strong heating decomposes CuSO₄ itself to black CuO + SO₂/SO₃, losing
extra mass and making x too BIG.
- Heat–cool–reweigh twice: the second, smaller loss checks you are approaching
constant mass. Incomplete dehydration is why x often comes out 4 rather than the true 5.
- Weigh only at room temperature — a hot tube sets up convection currents and the balance drifts.
- Dissolving the anhydrous solid in water re-forms the hydrate: exothermic, temperature rises,
ΔH negative — and the solution turns blue.
Q2 — manganese in three oxidation states
C = KMnO₄ (Mn +7, purple) · D = MnO₂ (Mn +4, black) · E = a Mn²⁺ salt (+2,
almost colourless).
- Test 1: H₂O₂ REDUCES purple MnO₄⁻ to colourless Mn²⁺ (H₂O₂ is the reducing agent; O₂ fizzes off).
- Test 2: MnO₂ CATALYSES 2H₂O₂ → 2H₂O + O₂ — vigorous effervescence; a glowing splint relights.
- Test 3: MnO₄⁻ oxidises I⁻ to I₂ — yellow-brown.
- Test 4: MnO₄⁻ + OH⁻ + MnO₂ → green manganate(VI), MnO₄²⁻; acid makes it disproportionate back
to purple/red-brown MnO₄⁻ + MnO₂.
- Tests 5 & 6: white ppt with AgNO₃ → Cl⁻; off-white Mn(OH)₂ ppt, insoluble in excess NaOH,
darkening in air → Mn²⁺. So E = MnCl₂.
- (e): the KMnO₄/ethanedioic-acid reaction is autocatalytic (Mn²⁺ catalyses it) — slow start,
fastest mid-way, slowing as the reactant runs out.
How the marking works (Version B)
- Write each answer as you would on the real paper, then press Mark my answer — an offline
rule-based marker scores it against the real marking points and tells you what it credited.
- Say "colourless", not "clear"; "white precipitate", not just "white"; name gases after testing them.
- Masses to 2 d.p. with units; table headers must say "boiling tube and contents" after
heating.