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Raoult's Law

The vapour pr...

Question

The vapour pressure of pure liquids 1 and 2 are $550 m m H g$ and $800 m m H g$ respectively at $350 K$ .
What will be the mole fraction of component 2 in liquid phase when total vapour pressure of the solution is $600 m m H g$ ?

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Solution

The correct option is B $0.2$
By Raoult's Law :
$P_{T} = p_{1}^{\circ} χ_{1} + p_{2}^{\circ} χ_{2}$
where
$P_{T}$ is the total pressure of the solution.
$p_{1}^{\circ}$ is the vapour pressure of pure component 1
$p_{2}^{\circ}$ is the vapour pressure of pure component 2
$χ_{1}$ is the mole fraction of component 1 in liquid phase.
$χ_{2}$ is the mole fraction of component 2 in liquid phase.

Putting the values:
$600 = 550 χ_{1} + 800 χ_{2} 600 = 550 χ_{1} + 800 (1 - χ_{1}) 600 = 800 - 250 χ_{1} χ_{1} = 0.8$
$χ_{2} = 1 - 0.8 = 0.2$

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Raoult's Law

Standard XII Chemistry

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The vapour pressure of pure liquids 1 and 2 are 550 mm Hg and 800 mm Hg respectively at 350 K. What will be the mole fraction of component 2 in liquid phase when total vapour pressure of the solution is 600 mm Hg?

The vapour pressure of pure liquids 1 and 2 are $550 m m H g$ and $800 m m H g$ respectively at $350 K$ .
What will be the mole fraction of component 2 in liquid phase when total vapour pressure of the solution is $600 m m H g$ ?