At $100^\circ\text{C}$, pure water boils, so its vapour pressure ($p^\circ$) is equal to the atmospheric pressure, which is $760 \text{ mm Hg}$. Given vapour pressure of the solution ($p_s$) = $732 \text{ mm Hg}$. Mass of solvent water ($w_1$) = $100 \text{ g} = 0.1 \text{ kg}$. Molar mass of water ($M_1$) = $18 \text{ g mol}^{-1}$.

According to the relative lowering of vapour pressure: $\frac{p^\circ - p_s}{p_s} = \frac{n_2}{n_1}$ $\frac{760 - 732}{732} = \frac{n_2}{100/18}$ $\frac{28}{732} = \frac{n_2}{5.555}$ $n_2 = \frac{28 \times 5.555}{732} \approx 0.2125 \text{ mol}$

Now, calculate the molality ($m$) of the solution: $m = \frac{n_2}{w_1 \text{ (in kg)}} = \frac{0.2125}{0.1} = 2.125 \text{ mol kg}^{-1}$

Elevation in boiling point ($\Delta T_b$) is given by: $\Delta T_b = K_b \times m = 0.52 \times 2.125 \approx 1.105^\circ\text{C}$

Therefore, the boiling point of the solution is: $T_b = 100^\circ\text{C} + 1.105^\circ\text{C} = 101.105^\circ\text{C} \approx 101^\circ\text{C}$.