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>>The vapour pressure of pure liquids A an

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The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively. at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.

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Given: Vapour pressure of pure liquid A, PoA=450mmofHg

Vapour pressure of pure liquid A, PoA=700mmofHg

Total vapour pressure,a Ptotal=600mmofHg

Use the formula of Raoult’s law$$

600=(450–700)XA+700

250XA=100

XA=

250

100

=0.4

Use formula

XB=1−XA

Substitute the values

we get, XB=1−0.4=0.6

use formula PA=PoA×XA=450×0.4=180mmofHg

PB=PoB×XB=700×0.6=420mmofHg

Now, in the vapour phase:

Mole fraction of liquid A=

180+420

180

=0.30

Mole fraction of liquid B,YB=1−YA=1–0.30=0.70