The boiling point of an aqueous solution containing $10 \mathrm{g}$of urea $(\text{ molar mass }=60)$ in $250 \mathrm{g}$ of water $\left({\mathrm{L}}_{\mathrm{v}}\right.$$\left.=539 \mathrm{cal} {\mathrm{g}}^{-}\right)$ will be

Relation between K_b and L_v (heat of vaporization):

 ${\mathrm{K}}_{\mathrm{b}}=\frac{\mathrm{R}{\left({\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}}\right)}^2}{1000{\mathrm{L}}_{\mathrm{v}}}$

${\mathrm{T}}_{\mathrm{b}}^{\circ }$ is the boiling point of pure solvent. ${\mathrm{T}}_{\mathrm{b}}^0=100+273=373 \mathrm{K}$

$\begin{array}{l}{\mathrm{K}}_{\mathrm{b}}=\frac{2 \mathrm{cal}/\mathrm{K}.\mathrm{mol} \times 373 \mathrm{K}\times 373 \mathrm{K}}{1000 \times 539 \mathrm{cal}/\mathrm{g}}\\ =0.516 \mathrm{K} \mathrm{g} {\mathrm{mol}}^{-1}\end{array}$

 ${\mathrm{\Delta T}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{b}}\times {\mathrm{W}}_2\left(\text{in }\mathrm{g}\right)}{{\mathrm{M}}_2\times {\mathrm{W}}_1\left(\text{in }\mathrm{kg}\right)}$

Substituting the given values:

$\begin{array}{r}∆{\mathrm{T}}_{\mathrm{b}}=\frac{0.516 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\times 10 \mathrm{g}}{60 \mathrm{g} {\mathrm{mol}}^{-}\times 0.25 \mathrm{kg}}=0.34 \mathrm{K}\\ \text{B}\mathrm{oiling} \mathrm{point}=373+0.34 \mathrm{K}=373.34 \mathrm{K}\end{array}$