Question

The vapour pressure of pure liquids A and B are 450 and 700mm Hg at 350 K

respectively. Find out the composition of the liquid mixture if total vapour

pressure is 600mm Hg. Also find the composition of the vapour phase.​

Answer 1

Answer:

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Explanation:

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=

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