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The vapour pressure of pure liquids A and B at 400 K are 450 and 700 mmHg respectively. Find out the composition of liquid mixture if total vapour pressure at this temperature is 600 mmHg.

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using formula of Raoult's law 600= (450-700) XA +700 250 XA =100 XA= 100÷250= 0.4 XB= 1- XA XB= 1-0.4=0.6 PA= PA0 × XA= 450× 0.4=180 mm of Hg PB = PB0× XB= 700×0.6= 420 mm of Hg Mole fraction of Liquid A= YA 180÷(180+120)= 0.30 Mole fraction of Liquid B= YB= 1-YA=...
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using formula of Raoult's law

600= (450-700) XA +700

250 XA =100

XA= 100÷250= 0.4

 

XB= 1- XA

XB= 1-0.4=0.6

PA= PA0 × XA= 450× 0.4=180 mm of Hg

PB = PB0× XB= 700×0.6= 420 mm of Hg

Mole fraction of Liquid A= YA 180÷(180+120)= 0.30

Mole fraction of Liquid B= YB= 1-YA= 1-0.30= 0.70

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