18 g of glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ is dissolved in 1 kg of water in a saucepan. The temperature at which water boil at 1.013 bar is $\left({\mathrm{K}}_{\mathrm{b}}\mathrm{for} \mathrm{water} \mathrm{is} 0.52 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\right)$

Wt. of glucose = $\mathrm{\omega }=18\mathrm{g}$

Wt. of water = W = 1 kg = 1000 g

GMW of Glucose $\left({\mathrm{C}}_6{\mathrm{H}}_1{\mathrm{O}}_6\right)=M=180 g$

B.P of pure solvent (Water) $={\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}}={100}^{\mathrm{o}}\mathrm{C}=373\mathrm{k}$

B.P of solution =${\mathrm{T}}_{\mathrm{b}}=?$

Molal elevation constant ${\mathrm{K}}_{\mathrm{b}}=0.52 \mathrm{k} {\mathrm{Kgmol}}^{-1}$

$∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}}={\mathrm{K}}_{\mathrm{b}}\times \frac{\mathrm{\omega }}{\mathrm{M}}\times \frac{1000}{{\mathrm{W}}_{\left(\mathrm{gr}\right)}}$

${\mathrm{T}}_{\mathrm{b}}-37.=0.52\times \frac{\cancel{18}}{{\cancel{180}}_{10}}\times \frac{\cancel{1000}}{\cancel{1000}}$

${\mathrm{T}}_{\mathrm{b}}-372=0.052$

${\mathrm{T}}_{\mathrm{b}}=373.052\mathrm{k}$