Question

The vapour pressure of pure liquids A and B are 400 and 600mmHg, respectively at 298K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquids B is 0.5 in the mixture. What will be the vapour pressure of the final solution, the mole fraction of components A and B in vapour phase, respectively?
(A) 500 mm Hg, 0.5,0.5
(B) 500 mm Hg, 0.4,0.6
(C) 500 mm Hg, 0.5,0.5
(D) 500 mm Hg, 0.4,0.6

Answer 1

Hint: Find the total pressure using the formula $P_{total}=X_A \cdot \dot{P}_{A}+ X_B \cdot \dot{P}_{B}$. Then we can find the mole fraction.
 Complete step by step answer:
Let us find the $P_{total}$ by using the formula $P_{total}=X_A \cdot \dot{P}_{A}+ X_B \cdot \dot{P}_{B}$
Where $\dot{P}_{A}$ is the is the partial pressures of A and $\dot{P}_{B}$ is the is the partial pressures of B.
And $X_A$ is the mole fraction of A and $X_B$ is the mole fraction of B
$P_{total}=0.5 \times 400+ 0.5\times 600$
By solving we get,
$P_{total}$=50 mm Hg
Now, we can find mole fraction of A in vapour,
$Y_A= \dfrac{P_A}{ P_{total}}= \dfrac{0.5 \times 400}{500}=0.4$
Additional information:
The pressure exerted by a vapor in equilibrium with the condensed phases at a given temperature in a closed system is called the vapour pressure.
$P_{total}=X_A \cdot \dot{P}_{A}+ X_B \cdot \dot{P}_{B}$ equation is also known as Dalton’s law of partial pressure.
Note: In these types of questions sometimes $X_B$ may not be given. We can find $X_B$ by $X_B=1-X_A$