$100 \, g$ 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 $475 \, torr$.

(N/A) Number of moles of liquid $A$,$n_{A} = \frac{100}{140} \, mol = 0.714 \, mol$.
Number of moles of liquid $B$,$n_{B} = \frac{1000}{180} \, mol = 5.556 \, mol$.
Mole fraction of $A$,$x_{A} = \frac{n_{A}}{n_{A} + n_{B}} = \frac{0.714}{0.714 + 5.556} = 0.114$.
Mole fraction of $B$,$x_{B} = 1 - 0.114 = 0.886$.
Vapour pressure of pure liquid $B$,$P_{B}^{o} = 500 \, torr$.
Vapour pressure of liquid $B$ in the solution,$p_{B} = P_{B}^{o} x_{B} = 500 \times 0.886 = 443 \, torr$.
Total vapour pressure of the solution,$p_{total} = 475 \, torr$.
Vapour pressure of liquid $A$ in the solution,$p_{A} = p_{total} - p_{B} = 475 - 443 = 32 \, torr$.
Using Raoult's law,$p_{A} = P_{A}^{o} x_{A}$.
$P_{A}^{o} = \frac{p_{A}}{x_{A}} = \frac{32}{0.114} = 280.7 \, torr$.
Thus,the vapour pressure of pure liquid $A$ is $280.7 \, torr$ and its vapour pressure in the solution is $32 \, torr$.