Question

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

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

Answer 1

Answer: (D)

The number of moles of water in $100g=\frac{100}{18}=5.555$
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 $1kg$ of water
$100g$ of water corresponds to $0.1kg$
$m=\frac{0.212}{0.1}=2.12$
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
$T_b=100+1.1=101.1^{\circ }C\simeq 101^{\circ }C$