# 100 g of liquid A (molar mass $140 g mol^{–1}$) was dissolved in 1000 g of liquid B (molar mass $180 g mol^{–1}$). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

$\begin{array}{1 1}32torr\\36torr\\42torr\\52torr\end{array}$

Number of moles of liquid A (solute)=$\large\frac{100g}{140g mol^{-1}}$
$\Rightarrow \large\frac{5}{7}$ mole
Number of moles of liquid A (solvent)=$\large\frac{1000g}{180g mol^{-1}}$
$\Rightarrow \large\frac{50}{9}$ mole
Mole fraction of A
$x_A=\large\frac{x_A}{x_A+x_B}$
$\quad=\large\frac{\Large\frac{5}{7}}{\Large\frac{5}{7}+\frac{50}{9}}$
$\quad=\large\frac{\Large\frac{5}{7}}{\Large\frac{395}{63}}$
$\quad=\large\frac{5}{7}\times \frac{63}{395}$
$\quad=\large\frac{45}{395}$
$\quad=0.114$
Mole fraction of B in the solution
$x_B=1-0.114=0.886$
Applying Raoult's law
$P_A=x_AP^0_A$
$\;\;\;\;\;=0.114\times P^0_A$
$P_B=x_BP_B^0$
$\;\;\;\;\;=0.886\times 500$
$\;\;\;\;\;=443torr$
$P_{total}=P_A+P_B$
$475=0.114P_A^{\large\circ}+443$
$P_A^0=\large\frac{475-443}{0.114}$
$\;\;\;\;\;=280.7torr$
$P_A=0.114\times 280.7$ torr
$\;\;\;\;\;=32$ torr