Solutions

Concept:Use the boiling point elevation to find van’t Hoff factor and then the degree of dissociation.Explanation:The dissociation: $\mathrm{B} \rightleftharpoons 2\mathrm{C} + 2\mathrm{D}$ gives 4 particles from 1 molecule of B.Van’t Hoff factor: $i = 1 + (4-1)\alpha = 1 + 3\alpha$.Given $0.25\%$ (mass/mass) B: in 100 g solution, 0.25 g B and 99.75 g solvent S.Molar mass of B is $10$ times that of S: $M_{\mathrm{B}} = 10 M_{\mathrm{S}}$.Moles of B = $\frac{0.25}{10 M_{\mathrm{S}}} = \frac{0.025}{M_{\mathrm{S}}}$.Mass of solvent = $99.75\ \mathrm{g} = 0.09975\ \mathrm{kg}$.Molality $m = \frac{0.025/M_{\mathrm{S}}}{0.09975} = \frac{1000}{399 M_{\mathrm{S}}}$ (since $\frac{0.025}{0.09975} = \frac{1000}{399}$).Boiling point elevation: $\Delta T_b = 408 - 400 = 8\ \mathrm{K}$.$K_b = \frac{R (T_b^{\circ})^2 M_{\mathrm{S}}}{1000\,\Delta H_{\mathrm{vap}}}$ with $T_b^{\circ}=400\ \mathrm{K}$, $\Delta H_{\mathrm{vap}}=10R$.$K_b = \frac{R \times 160000 \times M_{\mathrm{S}}}{1000 \times 10R} = 16 M_{\mathrm{S}}$.$\Delta T_b = i K_b m$ gives $8 = (1+3\alpha) \times 16 M_{\mathrm{S}} \times \frac{1000}{399 M_{\mathrm{S}}}$.Simplify: $8 = (1+3\alpha) \times \frac{16000}{399}$ → $1+3\alpha = \frac{8 \times 399}{16000} = \frac{3192}{16000} = 1.995$.Thus $3\alpha = 0.995$ → $\alpha = 0.331667$.Mole percent of B dissociated = $33.1667\%$.Answer:$33.16\%$ (approximately $33.00\%$ as per given option).