The boiling point of water $(100° \mathrm{C})$ becomes $100.52°\mathrm{C},$ if $3$ grams of a nonvolatile solute is dissolved in $200\mathrm{ml}$ of water. The molecular weight of solute is (${\mathrm{K}}_{\mathrm{b}}$ forwateris $0.6K-kg mo{l}^{-1}$ ) 

First boiling point of water $= 100°\mathrm{C},$

 Final boiling point of water $= 100.52°$

$\begin{array}{l}\mathrm{w}\left(\mathrm{solute}\right)=3\mathrm{g},\mathrm{W}\left(\mathrm{solvent}\right)=200\mathrm{g},{\mathrm{K}}_{\mathrm{b}}=0.6\mathrm{K}-{\mathrm{kgmol}}^{-1}\\ {\mathrm{\Delta T}}_{\mathrm{b}}=100.52-100=0{.52}^{\circ }\mathrm{C}\end{array}$

$\begin{array}{l}m=\frac{K_b\times w\times 1000}{\mathrm{\Delta }{T}_b\times W}\\ =\frac{0.6\times 3\times 1000}{0.52\times 200}=\frac{1800}{104}=17.3{\mathrm{gmol}}^{-1}.\end{array}$