Question

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

Answer 1

Answer:

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= 100/250

=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=180 mm of Hg

PB=PoB× XB=700×0.6=420 mm of Hg

Now, in the vapour phase:

Mole fraction of liquid A= 180/180+420

 =0.30

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