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

Answer: (C)

$P_{T}=P^{\circ}_{X} x_{X} + P^{\circ}_{Y} x_{Y}$

where, $P_{T}=$ Total pressure

$P^{\circ} x=$ Vapour pressure of $X$ in pure state

$P^{\circ}{ }_{Y}=$ Vapour pressure of $Y$ in pure state

$x_{X}=$ Mole fraction of $X=1 / 4$

$x_{Y}=$ Mole fraction of $Y=3 / 4$

(i) When $T=300\, K , P_{T}=550\, mm \,Hg$

$\therefore 550=P_{X}^{\circ}\left(\frac{1}{4}\right)+P_{Y}^{\circ}\left(\frac{3}{4}\right)$

$\Rightarrow P_{X}^{\circ}+3 P_{Y}^{\circ}=2200$ ...(1)

(ii) When at $T=300\, K , 1$ mole of $Y$ is added,

$P_{T}=(550+10) \,mm\, Hg$

$\therefore x_{X}=1 / 5$ and $x_{Y}=4 / 5$

$\Rightarrow 560=P_{X}^{\circ}\left(\frac{1}{5}\right)+P_{Y}^{\circ}\left(\frac{4}{5}\right)$

or $P_{X}^{\circ}+4 P_{Y}^{\circ}=2800$ ...(2)

On solving equations (1) and (2), we get

$P_{Y}^{\circ}=600 \,mm\, Hg$ and $P_{X}^{\circ}=400 \,mm \,Hg$