Consider a binary mixture of two ideal volatile liquids A and B having vapour pressure of $400 m m H g$ and $600 m m H g$ in their pure state.
What will be the mole fraction of component A in vapour phase $(y_{A})$ if the total pressure $(P_{T})$ of the solution is $500 m m H g$ ?

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Solution

$χ_{A} = \frac{y_{A} \times P_{T}}{p_{A}^{\circ}}$
$χ_{A}$ is the mole fraction of component A in liquid phase

$χ_{B} = \frac{y_{B} \times P_{T}}{p_{B}^{\circ}}$
$χ_{B}$ is the mole fraction of component B in liquid phase

Also $χ_{A} + χ_{B} = 1$

So, $\frac{1}{P_{T}} = \frac{y_{A}}{p_{A}^{\circ}} + \frac{y_{B}}{p_{B}^{\circ}} \frac{1}{P_{T}} = \frac{y_{A}}{p_{A}^{\circ}} + \frac{1 - y_{A}}{p_{B}^{\circ}}$
Solving,

$P_{T} = \frac{p_{A}^{\circ} \times p_{B}^{\circ}}{p_{A}^{\circ} + (p_{B}^{\circ} - p_{A}^{\circ}) \times y_{A}}$
Here ,
$P_{T} = 500 m m H g p_{A}^{\circ} = 400 m m H g p_{B}^{\circ} = 600 m m H g$
$500 = \frac{400 \times 600}{400 + (600 - 400) \times y_{A}} 5 = \frac{24}{4 + 2 y_{A}} 20 + 10 y_{A} = 24 y_{A} = 0.4$
(d) is the correct oprion

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Consider a binary mixture of two ideal volatile liquids A and B having vapour pressure of 400 mm Hg and 600 mm Hg in their pure state. What will be the mole fraction of component A in vapour phase (yA) if the total pressure (PT) of the solution is 500 mm Hg?

Consider a binary mixture of two ideal volatile liquids A and B having vapour pressure of $400 m m H g$ and $600 m m H g$ in their pure state.
What will be the mole fraction of component A in vapour phase $(y_{A})$ if the total pressure $(P_{T})$ of the solution is $500 m m H g$ ?