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Q.

A solution is prepared by mixing 8.5g of ${CH}_{2} {Cl}_{2}$ and 11.9g ${CHCl}_{3} .$ If vapour pressure of ${CH}_{2} {Cl}_{2} & {CHCl}_{3}$ at 298K are 415 & 200 mm of Hg respectively. The mole fraction of in vapour phase is

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a

0.162

b

0.675

c

0.325

d

0.486

answer is C.

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Detailed Solution

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$\begin{array}{l} moles of {CH}_{2} {Cl}_{2} = \frac{8 .5}{85} = 0 .1 \\ moles of {CHCl}_{3} = \frac{11 .9}{119 .5} = 0 .1 \\ Pressure = P_{T} = P_{{CH}_{2} {Cl}_{2}}^{0} x_{{CH}_{2} {Cl}_{2}} + P_{{CHCl}_{3}}^{0} x_{{CHCl}_{3}} \\ = 415 \times \frac{0 .1}{0 .2} + 200 \times \frac{0 .1}{0 .2} \Rightarrow 307 .5 \\ Y_{{CHCl}_{3}} = \frac{p_{{CHCl}_{3}}}{P_{T}} = P_{{CHCl}_{3}}^{0} x_{{CHCl}_{3}} = \frac{100}{307 .5} = 0 .325 \end{array}$

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