$\pu{1 L}$ 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.


1 Answer 1


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.


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