A solution of urea (mol. mass 56 g mol-1) 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 kg mol-¹ respectively, the above solution will freeze at 

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}.\mathrm{m} \\ {\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}.\mathrm{m} \end{gathered}$$

if the value of m is same, then
Hence,  $\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}}=\frac{{\mathrm{\Delta T}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{b}}}$
or ${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{\Delta T}}_{\mathrm{b}}\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{b}}}$
$\left[{\mathrm{\Delta T}}_{\mathrm{b}}=100.18-100=0{.18}^{\circ }\mathrm{C}\right]$
$∆{\mathrm{T}}_{\mathrm{f}}=0.18\times \frac{1.86}{0.512}=0{.65}^{\circ }\mathrm{C}$
As the Freezing Point of pure water is 0°C, 
$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{T}}_1(\mathrm{temperature} \mathrm{of} \mathrm{pure} \mathrm{solvent})-{\mathrm{T}}_2 (\mathrm{when} \mathrm{solute} \mathrm{is} \mathrm{added}) \\ 0.654°\mathrm{C}=0-{\mathrm{T}}_2 \\ {\mathrm{T}}_2=-0.654°\mathrm{C} \end{gathered}$$