At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mm Hg, respectively. If in a mixture of a and B, the vapoure is 300 mm Hg, the mole fractions of a in the vapour phase, respectively, are -A. `0.6`B. `0.5`C. `0.8`D. `0.4`

At 300 K the vapour pressure of two pure liquids, A and B are 100 and

1.	At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mm Hg, respectively. If in a mixture of a and B, the vapoure is 300 mm Hg, the mole fractions of a in the vapour phase, respectively, are -A. `0.6`B. `0.5`C. `0.8`D. `0.4`
Answer» Correct Answer - D In equimolar liquid mixture `x_(A)=0.5, X_(B)=0.5` So, `P=0.5xx150+0.5xx100=125` Now let `y_(B)` be the mole fraction of vapour B then `y_(B)=(x_(B)p_(B)^(@))/(P)=(0.5xx100)/(125)=0.4`.

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At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mm Hg, respectively. If in a mixture of a and B, the vapoure is 300 mm Hg, the mole fractions of a in the vapour phase, respectively, are -A. `0.6`B. `0.5`C. `0.8`D. `0.4`

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