Chemistry, asked by mahimaverma2005, 1 month ago

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

Answers

Answered by chawlakanishka25
0

Answer:

sorry I don't know the ans

Answered by brainlylegend28
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

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