Question

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

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

Answer 1

Answer: (C)

$RLVP=\frac{\left(p^{\circ }-p\right)}{p^{\circ }}$
$[\because p^{\circ }=$ vapour pressure of $H_{2}O(g)$ at $373K=760bar$
Let, $d$ water $=1g/mL$
$\Rightarrow 54mL=54g$ water
$\because \chi \to$ mole fraction
$n\to$ 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