Question 12

Calculate the ebullioscopic constant ($K_b$) for solvent $S$: $K_b = \frac{R (T_b^0)^2 M_S}{1000 \Delta H_{vap}} = \frac{R (400)^2 M_S}{1000 \times 10R} = \frac{160000 M_S}{10000} = 16 M_S$ K kg mol$^{-1}$.

Determine molality ($m$) of $B$: Concentration = $0.25\%$ (w/w) $\implies$ $0.25$ g of $B$ in $99.75$ g of $S$. $m = \frac{n_B}{\text{mass of solvent in kg}} = \frac{0.25 / M_B}{99.75 / 1000} = \frac{0.25 / (10 M_S)}{0.09975} = \frac{0.025}{0.09975 M_S} \approx \frac{0.25}{M_S}$ mol/kg (using approximation $99.75 \approx 100$).

Use elevation in boiling point formula: $\Delta T_b = i \cdot K_b \cdot m$ $408 - 400 = i \cdot (16 M_S) \cdot \frac{0.25}{M_S}$ $8 = i \cdot 4 \implies i = 2$.

Calculate degree of dissociation ($\alpha$): $B \rightleftharpoons 2C + 2D$ Initial: 1, 0, 0 Equilibrium: $1-\alpha, 2\alpha, 2\alpha$ $i = 1 - \alpha + 2\alpha + 2\alpha = 1 + 3\alpha$ $2 = 1 + 3\alpha \implies 3\alpha = 1 \implies \alpha = 1/3$.

Mole percent of $B$ dissociated = $\alpha \times 100 = 33.33\%$.