At $100°\mathrm{C}$ the vapour pressure of a solution of 6.5 gm of a solute in 100 gm of water is 732 mm. If ${\mathrm{K}}_{\mathrm{b}}=0.52$, the boiling point of this solution will be ${}^{\circ }\mathrm{C}$ (nearest integer)

The volume of water in moles $100 \mathrm{g}=\frac{100}{18}=5.555$
Let n represent the solute's molecular weight.
The mole fraction of solute equals the relative decrease in vapour pressure.
$\frac{{\mathrm{P}}^0-\mathrm{P}}{{\mathrm{P}}^0}=\mathrm{X}$

$\frac{760-732}{760}=\frac{\mathrm{n}}{\mathrm{n}+5.555}$

$0.0368=\frac{\mathrm{n}}{\mathrm{n}+5.555}$

27.144 = n+5.555

$\mathrm{n}=\frac{5.555}{26.14}=0.212$

The number of moles of solute in 1 kg of water is the molality of the solution.

100 g of water equals 0.1 kg.

$\mathrm{m}=\frac{0.212}{0.1}=2.12$

The elevation in the boiling point is

$\Delta {\mathrm{T}}_{\mathrm{b}}={\mathrm{k}}_{\mathrm{b}}\mathrm{m}=0.52\times 2.12=1.1°\mathrm{C}$

The boiling point of the solution is

${\mathrm{T}}_{\mathrm{b}}=100+1.1=101.1°\mathrm{C}\simeq 101°\mathrm{C}$

Hence, the correct answer is 101⁰C.