# Boiling point elevation and liquid solutions

$$\pu{1 L}$$ of an aqueous solution of urea having density $$\pu{1.06 g mL-1}$$ is found to have elevation in boiling point $$\Delta T_\mathrm{b} = \pu{0.5 °C}.$$ If the temperature of this solution is increased to $$\pu{101.5 °C},$$ then calculate the amount of water which must have vaporized by doing so. Ebullioscopic constant of water is $$K_\mathrm{b} = \pu{0.5 K kg mol-1}.$$

I tried solving this question using the Clausius-Clapeyron equation and the basic ideal gas equation. I calculated the change in the number of (gaseous) moles of water that would occur by raising the solution's temperature to $$\pu{101.5 °C},$$ and then multiplied it by its molar mass, hoping to get the mass of water thus evaporated.

$$\Delta T = K_\mathrm{b} m$$
First solve for $$m_\mathrm{init}$$, the initial molality of urea. Second figure out at what molality $$m_\mathrm{fin}$$ the boiling point is elevated by $$\Delta T = \pu{1.5 °C}.$$ Since the amount of urea is constant, the % change in the mass of water is then given by
$$\% \left(\frac{\Delta w_{H2O}}{w_{H2O,\mathrm{init}}}\right)= \frac{m_\mathrm{init}^{-1}-m_\mathrm{fin}^{-1}}{m_\mathrm{init}^{-1}}$$
where $$m$$ is the molality of urea, $$w$$ mass solvent.