Question

The vapour pressure of $H_{2}O$ is $23.8\text{mm Hg}$ at ${25}^{o}C$. What is the vapour pressure of a solution of $28.5\text{g}$ of sucrose, $C_{12}{H}_{22}{O}_{11}\left(\text{Molecular mass= 342 u}\right)$ in $100\text{g}$ of $H_{2}O$?

• A. $23.1\text{mm Hg}$
• B. $32.5\text{mm Hg}$
• C. $20.6\text{mm Hg}$
• D. $20.00\text{mm Hg}$

Answer 1

The correct option is A $23.1\text{mm Hg}$

$\text{Mole of sucrose}=\frac{28.5}{342}=0.083\text{mol}$
$\text{Moles of}{H}_{2}O=\frac{100}{18}=5.55\text{mol}$
$\therefore \text{mole fraction of solute}=\frac{0.083}{0.083+5.55}=0.0148$
By using raoult's law,
$\frac{p^0-p}{p^0}=x_B...\left(i\right)$
Where, $x_B=\text{Mole fration of solute}{p}^0=\text{Vapour pressure of water}p=\text{Vapour pressure of solution}$
$\frac{23.8-p}{23.8}=0.0148$
$p=23.404\text{mm Hg}$.