Question

Hey experts. Answer this in detail.

Answer 1

Hi

Pls refer s chand book of Chemistry for this.....

Answer 2

Answer:

Explanation:

Solution

Given that

Vapour pressure of pure liquid A, PoA= 450 mm of Hg

Vapour pressure of pure liquid A, PoA= 700 mm of Hg

Total vapour pressure, ptotal = 600 mm of Hg

Use the formula of Raoult’s law

 

600 = (450 – 700) XA  + 700

250 XA = 100

XA = 100/250 = 0.4

Use formula

XB = 1 - XA

Plug 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

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