At $100° \mathrm{C}$ the vapour pressure of a solution of $6.5\mathrm{g}$ of a solute in $100 \mathrm{g}$water is $732 \mathrm{mm}.$ If$\mathrm{Kb} = 0.52$, the boiling point of this solution will be 

$\left(\frac{P^0-P_s}{P^0}\right)=\frac{n}{N}=\frac{w_{\text{solute }}}{M_{\text{solute }}}\times \frac{M_{\text{solvent }}}{W_{\text{solvent }}}$

at ${100}^{\circ }\mathrm{C},P^0=760\mathrm{mm}$

$\begin{array}{l}\frac{760-732}{760}=\frac{6.5\times 18}{M_{\text{solute }}\times 100}\\ M_{\text{solute }}=31.75{\mathrm{gmol}}^{-1}\\ \mathrm{\Delta }{T}_b=m\times K_b=\frac{w_{\text{solute }}\times 1000}{M_{\text{solute }}\times w_{\text{solvent }}}\times K_b\\ \mathrm{\Delta }{T}_b=\frac{0.52\times 6.5\times 1000}{31.75\times 100}={1.06}^0\mathrm{C}\end{array}$

$\therefore$boiling point of solution $={100}^0\mathrm{C}+{1.06}^0\mathrm{C}\simeq {101}^0\mathrm{C}$