Question

Two liquids X and Y forms an ideal solution. At 300 K, the vapour pressure of the solution containing 1 mole of X and 3 moles of Y is 550 mm Hg. At the same temperature if 1 mole of Y is further added to this solution, the vapour pressure of the solution increases by 10 mm Hg.
The 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 400 and 600

$x_X=\frac{1}{4}=0.25,x_Y=0.75P_{\text{total}}=x_X\times P_x^0+x_Y\times P_Y^{0}i.e.550=0.25P_x^0+0.75P_Y^{0}or,2200=P_x^0+3P_Y^0\to \left(1\right)$
After adding 1 mole of Y,
$x_X=\frac{1}{5}=0.20,x_Y=0.80\therefore 560=0.20P_X^0+0.80P_Y^{0}or,2800=P_X^0+4P_Y^0\to \left(2\right)$
Solving the two equations, we get
$P_Y^0=600$ mm Hg
$P_X^0=400$ mm Hg