Chemistry — Solid State, Solutions, Electrochemistry, Kinetics & Organic — Practice

Question 1

What is the freezing point depression of 2.0 g of benzoic acid (M = 122 g/mol) dissolved in 25 g of benzene? Kf for benzene = 5.12 K kg mol⁻¹.

Accepted Answer

Molality: m = (wB × 1000)/(MB × wA) = (2.0 × 1000)/(122 × 25) = 2000/3050 = 0.656 mol/kg. ΔTf = Kf × m = 5.12 × 0.656 = 3.36 K. Note: Benzoic acid dimerises in benzene (non-polar solvent), so the observed ΔTf would be ~HALF this value, giving i ≈ 0.5. But if no mention of dimerisation, use the formula directly: ΔTf = 3.36 K.

Question 2

Calculate the EMF of the cell: Zn | Zn²⁺(0.001 M) || Cu²⁺(0.1 M) | Cu at 298 K. E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.

Accepted Answer

E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V. n = 2 (two electrons transferred). Q = [Zn²⁺]/[Cu²⁺] = 0.001/0.1 = 0.01. Nernst: E = E° − (0.0591/n) log Q = 1.10 − (0.0591/2) log(0.01) = 1.10 − (0.02955)(−2) = 1.10 + 0.059 = 1.159 V ≈ 1.16 V.

Question 3

Give the IUPAC name of the product formed when aniline is treated with (i) Bromine water, (ii) Acetic anhydride, (iii) NaNO₂ + HCl at 0-5°C followed by CuBr.

Accepted Answer

(i) Aniline + Br₂ water → 2,4,6-TRIBROMOANILINE (white precipitate, ring highly activated at o- and p- positions by -NH₂ group). (ii) Aniline + (CH₃CO)₂O → ACETANILIDE (N-phenylethanamide, CH₃CONHC₆H₅) + CH₃COOH. This is acetylation of the amino group. (iii) Aniline + NaNO₂ + HCl → BENZENEDIAZONIUM CHLORIDE (0-5°C), then + CuBr (Sandmeyer reaction) → BROMOBENZENE + N₂.

Practice — Chemistry — Solid State, Solutions, Electrochemistry, Kinetics & Organic

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