Q. Two liquids X and Y form an ideal solution at 300K, vapour pressure of the solution containing 1 mol of X and 3, mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg)of Xand Y in their pure states will be, respectively
A) 200 and 300
B) 300 and 400
C) 400 and 600
D) 500 and 600
Explanation
(c)
$P_{T} = P_{A}^{\circ}\ X_{A} + p_{B}^{\circ}X_{B}$
$550 = P_{A}^{\circ} \times \frac{1}{4} + \ p_{B}^{\circ} \times \frac{3}{4}$
Thus, $\ \ \ \ \ \ \ \ \ \ P_{A}^{\circ} + 3p_{B}^{\circ} = 2200\ \$ …(i)
When, 1 mol of Y is further added to the solution
$560 = \ P_{A}^{\circ}\ \times \frac{1}{5} + \ p_{B}^{\circ} \times \frac{4}{5}\$
Thus, $P_{A}^{\circ} + 4p_{B}^{\circ} = 2800\$ …(ii)
On subtraction II-I
$p_{B}^{{^\circ}}$=2800-2200
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p_{B}^{\circ}$ = 600
Putting the value of $p_{B}^{\circ}$ in Eq. (i)
$P_{A}^{\circ} + 3 \times 600 = 2200$
$P_{A}^{\circ}$ = 2200-1800=400.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.