Two liquids A and B have, PA0 and PB0=1:3 at a certain temperature. Assume A and B form an ideal solution, and the ratio of mole of A to B in the liquid phase are 1:3 , then mole fraction of A in vapour phase in equilibrium with the solution -

Question

Two liquids A and B have, PA0 and PB0=1:3 at a certain temperature. Assume A and B form an ideal solution, and the ratio of mole of A to B in the liquid phase are 1:3 , then mole fraction of A in vapour phase in equilibrium with the solution - (A) 0.1 (B) 0.2 (C) 0.5 (D) 1.0

Tardigrade Admin · Accepted Answer

Pt=PA0xA+PB0xB=(1/4)PA0+(3/4)× 3PA0 =(10/4)PA0=(5/2)PA0 PT⋅ YA=PA0xA⇒ yA=(PA0 x (1/4)/(5)2 PA0)=(1/10)

Q. Two liquids $A$ and $B$ have, $P_{A}^{0}$ and $P_{B}^{0} = 1 : 3$ at a certain temperature. Assume $A$ and $B$ form an ideal solution, and the ratio of mole of $A$ to $B$ in the liquid phase are $1 : 3$ , then mole fraction of A in vapour phase in equilibrium with the solution -

$0.1$

$0.2$

$0.5$

$1.0$

$Pt = P_{A}^{0} x_{A} + P_{B}^{0} x_{B} = \frac{1}{4} P_{A}^{0} + \frac{3}{4} \times 3 P_{A}^{0}$
$= \frac{10}{4} P_{A}^{0} = \frac{5}{2} P_{A}^{0}$
$P_{T} \cdot Y_{A} = P_{A}^{0} x_{A} \Rightarrow y_{A} = \frac{P _{A}^{0} x \frac{1}{4}}{\frac{5}{2} P _{A}^{0}} = \frac{1}{10}$

Q. Two liquids A and B have, PA0​ and PB0​=1:3 at a certain temperature. Assume A and B form an ideal solution, and the ratio of mole of A to B in the liquid phase are 1:3 , then mole fraction of A in vapour phase in equilibrium with the solution -

0.1

0.2

0.5

1.0