Question

1) the v.p of pure liquid A&B are 450 mmHg & 700mmHg resp. at 350k.​

Answer 1

It is given that:

PAo = 450 mm of Hg

PBo = 700 mm of Hg

ptotal = 600 mm of Hg

From Raoult's law, we have:

ptotal = PA + PB

Therefore, xB = 1 - xA

= 1 - 0.4

= 0.6

Now, PA = PAo xA

= 450 × 0.4

= 180 mm of Hg

and PB = PBo xB

= 700 × 0.6

= 420 mm of Hg

Now, in the vapour phase: Mole fraction of liquid A = PA / (PA + PB )

=180 / (180+420)

= 180/600

= 0.30

And, mole fraction of liquid B = 1 - 0.30

= 0.70