Two liquids $X$ and $Y$ form an ideal solution. At $300 \mathrm{~K}$, vapour pressure of the solution…
Two liquids $X$ and $Y$ form an ideal solution. At $300 \mathrm{~K}$, vapour pressure of the solution containing 1 mol of $X$ and $3 \mathrm{~mol}$ of $Y$ is $550 \mathrm{~mm} \mathrm{Hg}$. At the same temperature, if $1 \mathrm{~mol}$ of $Y$ is further added to this solution, vapour pressure of the solution increases by $10 \mathrm{~mm} \mathrm{Hg}$. Vapour pressure (in $\mathrm{mmHg}$ ) of $X$ and $Y$ in their pure states will be, respectively :
200 and 300
300 and 400
400 and 600
500 and 600
Solution
$
\begin{aligned}
& P_T=P_X^0 x_X+P_Y^0 x_Y \\
& x_X=\text { mol fraction of } X \\
& x_Y=\text { mol fraction of } Y \\
& \therefore 550=P_x^{\circ}\left(\frac{1}{1+3}\right)+P_Y^{\circ}\left(\frac{3}{1+3}\right) \\
& =\frac{P_X^0}{4}+\frac{3 P_Y^0}{4} \\
& \therefore 550(4)=P_X^0+3 P_Y^0
\end{aligned}
$
Further $1 \mathrm{~mol}$ of $Y$ is added and total pressure increases by $10 \mathrm{~mm} \mathrm{Hg}$.
$
\begin{aligned}
& \therefore 550+10=\mathrm{P}_X^{\circ}\left(\frac{1}{1+4}\right)+\mathrm{P}_Y^{\circ}\left(\frac{4}{1+4}\right) \\
& \therefore 560(5)=\mathrm{P}_{\mathrm{X}}^{\circ}+4 \mathrm{P}_Y^{\circ} \ldots \ldots \ldots \ldots(2)
\end{aligned}
$
By solving (1) and (2)
We get, $\mathrm{P}_{\mathrm{x}}^{\circ}=400 \mathrm{~mm} \mathrm{Hg}$
$
P_Y^{\circ}=600 \mathrm{~mm} \mathrm{Hg}
$