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Question

Liquids P and Q form an ideal solution. The vapour pressures of pure liquids P and Q at 80°C are 300 mm and 100 mm of Hg, respectively. Suppose that the vapour above a solution composed of 1.0 mole of P and 1.0 mole of Q at 80°C is collected and condensed. This condensate is then heated to 80°C and vapour is again condensed to form a liquid R. What is the mole fraction of P in R?

Options

A.0.5
B.0.75
C.0.90
D.0.10

Solution

$\left( \frac{n_{P}}{n_{Q}} \right)_{2^{\text{nd}}\text{ condense}} = \left( \frac{n_{P}}{n_{Q}} \right)_{\text{initial}}.\left( \frac{P_{P}^{o}}{P_{Q}^{o}} \right)^{n}$

Where n = number of condensation steps.

Now, $\left( \frac{n_{P}}{n_{Q}} \right)_{\text{final}} = \frac{1}{1} \times \left( \frac{300}{100} \right)^{2} = \frac{9}{1}$

$\therefore X_{P} = \frac{9}{9 + 1} = 0.90$

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