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`        Vapor pressure of A is 300 torr and that of B is 800 torr. Mole fraction of A in a solution is 0.6, if at constant temperature the pressure of the system is decreased, find mole fraction of B in vapour when first bubble is formed.`
9 months ago

## Answers : (1)

Arun
23513 Points
```							a)  pB0 = 300 torr pB0 = 800 torryA = pA/P = pA0xA/P where yA and yB are mole fraction of A and B in the vapour yB = pB0xA/P ∴ yA = yB = pB0xA/pB0xB0.025/0.75 = 300xA/800xB i.e.,  xA/xB = 8/9   i.e.,   xA = 8/17 = 0.47Composition of last drop: xB = 9/17 = 0.53 (B) Pressure of last drop: p = pA0xA + pB0xB  (c) At boiling point: p = 760       760o = 300 xA + 800 (1−xA) xA = 0.08,          ∴ xB = 0.92 (d) For last drop:      xA = 0.25, xB = 0.75 P = 300 × 0.25 + 800 × 0.75 = 675 torr. (e) yA = pA/P yA = 300×0.25/675 = 0.111 yB = 0.889 Here, yA and yB are composition of vapour of last drop.
```
9 months ago
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