Question

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

Answer 1

Explanation:

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

Vapour pressure of pure liquid A,

PoA=700mm of Hg

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)=180mmofHg

PB=PoB×XB

=(700×0.6)=420mmofHg

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