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. 

Am I doing something incorrectly? My answer does not seem to match the answer given in my book at all. If yes, please help me out here.
Use the ebullioscopic equation (the first equation) in this Wikipedia article,

$$\Delta T = K_\mathrm{b} m$$

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.