A solution of urea (mol. mass 56 g mol^-l) boils at 100.18 °C at the atmospheric pressure. If K_f and K_b for water are 1.86 and 0.512 K kg mol^-l respectively, the above solution will freeze at

$\begin{array}{l}\text{ As }{\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\cdot \mathrm{m}\\ {\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{K}}_{{\mathrm{b}}^{.}}\mathrm{m}\end{array}$
Hence, we have $\mathrm{m}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}}=\frac{{\mathrm{\Delta T}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{b}}}$
or ${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{\Delta T}}_{\mathrm{b}}\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{b}}}\Rightarrow \left[{\mathrm{\Delta T}}_{\mathrm{b}}=100.18-100=0{.18}^{\circ }\mathrm{C}\right]$
$\therefore {\mathrm{\Delta T}}_{\mathrm{f}}=0.18\times \frac{1.86}{0.512}=0{.654}^{\circ }\mathrm{C}$
As the freezing point of pure water is 0°C,
$\begin{array}{l}{\mathrm{\Delta T}}_{\mathrm{f}}=0-{\mathrm{T}}_{\mathrm{f}}\\ 0.654=0-{\mathrm{T}}_{\mathrm{f}}\\ \therefore {\mathrm{T}}_{\mathrm{f}}=-0.654\end{array}$
Thus the freezing point of solution will be $-{0.654}^{\circ }\mathrm{C}$.