$18 g$ of glucose ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ ( molar mass $180 \mathrm{g}$${\mathrm{mol}}^{-}\text{) }$ is dissolved in $1 \mathrm{kg}$ of water in a sauce pan. The temperature at which water will boil ($1.013$ bar pressure) $\left(K_b\right.$ for water $\left.=0.52 K \mathrm{kg} {\mathrm{mol}}^{-}\right)$ will be

Weight of solute, $W_2=18 \mathrm{g}$ 

Wt. of solvent, $W_1=1 kg$

$\mathrm{mol}.\mathrm{wt}.=180 \mathrm{g} {\mathrm{mol}}^{-} \mathrm{and} {\mathrm{K}}_{\mathrm{b}}=0.52 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-}$ 

It is known as:

$\begin{array}{l}{\mathrm{\Delta T}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{b}}\times {\mathrm{W}}_2\text{ in }\mathrm{g}}{{\mathrm{M}}_2\times {\mathrm{W}}_1\text{ in }\mathrm{kg}}\\ =\frac{0.52 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-}\times 18 \mathrm{g}}{180 \mathrm{g} {\mathrm{mol}}^{-}\times 1 \mathrm{kg}}=0.052 \mathrm{K}\\ \mathrm{T}-{\mathrm{T}}_{\circ }=∆{\mathrm{T}}_{\mathrm{b}}\\ \mathrm{T}-373 \mathrm{K}=0.052 \mathrm{K}\\ \mathrm{T}=373 \mathrm{K}+0.052 \mathrm{K}=373.052\mathrm{K}\end{array}$