Question

At 100^oC (boiling point)
Vapour pressure of water P^o = P_atm = 760 ml
$\therefore \frac{P^o-P_s}{P^o}=X_{solute}$
$\Rightarrow \frac{760-732}{760}=\frac{n_{solute}}{n_{solvent}}$
$\Rightarrow \frac{28}{760}=\frac{6.5/m}{100/18}$
$\Rightarrow m= \frac{6.5\times 18\times 760}{28\times 100}\approx 32$
Now,
$∆$T_b = K_b molality
= 0.52 $\times \frac{6.5/32}{0.1}$
$=\frac{0.52\times 6.5}{32\times 0.1}$
=1.05 $\approx$ 1s
$\therefore$ Boiling point of solution = 100 + 1 = 101^oC