Two liquids P and Q form an ideal solution at $300$ K, the vapour pressure of a solution of $1$ mole of P and x mole of Q is $550$ mm. If the vapour pressure of pure P and pure Q are $400$ mm and $600$ mm respectively then x is?

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Solution

The correct option is C $3$
Mole fraction of Q = $x_{Q} = \frac{x}{1 + x}$

Mole fraction of P = $x_{P} = \frac{1}{1 + x}$

Vapour pressure of pure P = $P_{P}^{o} = 400 m m$
Vapour pressure of pure Q = $P_{Q}^{o} = 600 m m$
Vapour pressure of solution = P = 550 mm

According to Raoult's law: $P = P_{P}^{o} x_{P} + P_{Q}^{o} x_{Q}$

Substitute all values:
$550 m m = (400 m m) \frac{1}{1 + x} + (600 m m) \frac{x}{1 + x}$
$550 = \frac{400 + 600 x}{1 + x}$
$550 + 550 x = 400 + 600 x$
$50 x = 150$
$x = 3$

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