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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.
4 months ago

a)  pB0 = 300 torr pB0 = 800 torr

yA = pA/P = pA0xA/P

where yA and yB are mole fraction of A and B in the vapour

yB = pB0xA/P

∴ yA = yB = pB0xA/pB0xB

0.025/0.75 = 300xA/800xB

i.e.,  xA/xB = 8/9

i.e.,   xA = 8/17 = 0.47

Composition 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.

4 months ago
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