Question

At ${100}^{\circ }C$ the vapour pressure of a solution of $6.5g$ of a solution in $100g$ water is $732mm$. If $K_b=0.52$, the boiling point of this solution will be:

• A. ${101}^{\circ }C$
• B. ${100}^{\circ }C$
• C. ${102}^{\circ }C$
• D. ${103}^{\circ }C$

Answer 1

The correct option is A ${101}^{\circ }C$

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

$\Delta T_b=k_{b}m=0.52\times 2.12={1.1}^{o}C$

The boiling point of the solution is

$T_b=100+1.1={101.1}^{o}C\simeq {101}^{o}C$

Hence, the correct option is $A$