Question

Two liquids $X$ and $Y$ form an ideal solution. At $300\,K$, vapour pressure of the solution containing $1 \,mol$ of $X$ and $3\,mol \,of\, Y$ is $550 \,mm\, Hg$. At the same temperature, if $1\, mol$ of $Y$ is further added to this solution, vapour pressure of the solution increases by $10\, mm \,Hg$. Vapour pressure ($in\, mm\,Hg$) of $X$ and $Y$ in their pure states will be, respectively :

• A. $200$ and $300$
• B. $300$ and $400$
• C. $400$ and $600$
• D. $500$ and $600$

Answer 1

The Correct Option is C

$P_{T}=P^{o}_{x}x_{x}+P^{o}_{y}x_{y}$ $x_{x}=mol$ fraction of $X$ $x_{y}=mol$ fraction of $Y$ $\therefore 550=P^{o}_{x}\left(\frac{1}{1+3}\right)+P^{o}_{y}\left(\frac{3}{1+3}\right)$ $=\frac{P^{o}_{x}}{4}+\frac{3P^{o}_{y}}{4} \therefore 550\left(4\right)=P^{o}_{x}+3P^{o}_{y} \,............\left(1\right)$ Further 1 mol of Y is added and total pressure increases by 10 mm Hg. $\therefore 550+10=P^{o}_{x}\left(\frac{1}{1+4}\right)+P^{o}_{y} \left(\frac{4}{1+4}\right)$ $\therefore 560\left(5\right)=P^{o}_{x}+4P^{o}_{y} \,.............\left(2\right)$ By solving $\left(1\right)$ and $\left(2\right)$ We get, $P^{o}_{x}= 400\, mm \,Hg$ $P^{o}_{y}=600\, mm \,Hg$