Question

$160\, g$ of non-volatile solute '$A$' is dissolved in $54\, ml$ of water at $373\, K$. What is the vapor pressure of aqueous solution of A. (given molecular weight of $A =160\, g \cdot mol ^{-1}$ )

• A. 760 Torr
• B. 720 Torr
• C. 570 Torr
• D. 450 Torr

Answer 1

Answer: (C)

$R L V P=\frac{\left(p^{\circ}-p\right)}{p^{\circ}}$
$\left[\because p^{\circ}=\right.$ vapour pressure of $H_{2} O(g)$ at $373\, K=760\, bar$
Let, $d$ water $=1\, g / m L$
$\Rightarrow 54\, m L=54\, g$ water
$\because \chi \rightarrow$ mole fraction
$n \rightarrow$ mole $]$
$\Rightarrow \chi_{A}=\frac{n_{A}}{n_{A}+n_{H_{2} O}}$
$\Rightarrow \frac{760-p}{760}=\frac{\frac{160}{160}}{\frac{160}{160}+\frac{54}{18}}=\frac{1}{4}$
$\Rightarrow p=\frac{3}{4} \times 760=570$ tour