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JEE Advanced 2024 Paper 1 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

Match the aqueous solution in List-I with its freezing-point depression Δ Tf in List-II. Take Kf=1.86 K kg mol^-1 and assume ideal behaviour, and…

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Match the aqueous solution in List-I with its freezing-point depression ΔTf\Delta T_f in List-II. Take Kf=1.86 K kg mol−1K_f=1.86\ \mathrm{K\,kg\,mol^{-1}} and assume ideal behaviour, and complete dissociation of strong electrolytes.

| List-I | List-II (ΔTf\Delta T_f in K) |
|---|---|
| P. 0.10 m0.10\ m AlCl3\mathrm{AlCl_3} | 1. 0.2050.205 |
| Q. 0.10 m0.10\ m weak monobasic acid HA, 10% dissociated | 2. 0.4650.465 |
| R. 0.10 m0.10\ m NaCl plus 0.10 m0.10\ m glucose | 3. 0.5580.558 |
| S. 0.050 m0.050\ m K4[Fe(CN)6]\mathrm{K_4[Fe(CN)_6]} | 4. 0.7440.744 |
| | 5. 0.9300.930 |
  1. A. P→4, Q→1, R→3, S→2
  2. B. P→4, Q→1, R→2, S→3
  3. C. P→5, Q→1, R→3, S→2
  4. D. P→4, Q→3, R→1, S→5
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Answer: (A) P→4, Q→1, R→3, S→2

ΔTf=Kf×(total solute particles, molal)\Delta T_f=K_f\times(\text{total solute particles, molal}). P: AlCl3→Al3++3Cl−\mathrm{AlCl_3\to Al^{3+}+3Cl^-}, i=4i=4, 0.40 m0.40\ m particles, so ΔTf=1.86×0.40=0.744\Delta T_f=1.86\times0.40=0.744 K: 4. Q: i=1+α=1.1i=1+\alpha=1.1, particles 0.11 m0.11\ m, ΔTf=0.2046≈0.205\Delta T_f=0.2046\approx0.205 K: 1. R: NaCl gives 0.20 m0.20\ m and glucose 0.10 m0.10\ m, total 0.30 m0.30\ m, ΔTf=0.558\Delta T_f=0.558 K: 3. S: K4[Fe(CN)6]\mathrm{K_4[Fe(CN)_6]} gives 5 ions, 0.050×5=0.25 m0.050\times5=0.25\ m, ΔTf=0.465\Delta T_f=0.465 K: 2. Value 5 (0.9300.930 K, 0.5 m0.5\ m particles) is not matched. The code is P→4, Q→1, R→3, S→2 (checked numerically).
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