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Q.
At $100^{\circ} C$ the vapour pressure of a solution of $6.5 \,g$ of a solute in $100 \,g$ water is $732\, mm$. If $K_b = 0.52$, the boiling point of this solution will be:
The number of moles of water in $100 g =\frac{100}{18}=5.555$
Let $n$ be the number of moles of solute.
The relative lowering in the vapour pressure is equal to the mole fraction of solute
$\frac{ P ^{0}- P }{ P ^{0}}= X$
$
\frac{760-732}{760}=\frac{n}{n+5.555}
$
$
0.0368=\frac{n}{n+5.555}
$
$
27.144= n +5.555
$
$
n =\frac{5.555}{26.14}=0.212
$
The molality of the solution is the number of moles of solute in $1 kg$ of water
$100 g$ of water corresponds to $0.1 kg$
$
m =\frac{0.212}{0.1}=2.12
$
The elevation in the boiling point is
$
\Delta T _{ b }= k _{ b } m =0.52 \times 2.12=1.1^{\circ} C
$
The boiling point of the solution is
$
T _{ b }=100+1.1=101.1^{\circ} C \simeq 101^{\circ} C
$