Question 8

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

Answer

It is given that:

P_{A}^{o} = 450 mm of Hg

P_{B}^{o }= 700 mm of Hg

p_{total }= 600 mm of Hg

From Raoult's law, we have:

p_{total }= P_{A }+ P_{B}

Therefore, *x*_{B} = 1 - x_{A}

= 1 - 0.4

= 0.6

Now, P_{A}^{ }= P_{A}^{o }x_{A}

= 450 × 0.4

= 180 mm of Hg

and P_{B}^{ }= P_{B}^{o }x_{B }

= 700 × 0.6

= 420 mm of Hg

Now, in the vapour phase: Mole fraction of liquid A = P_{A }**/ **(P_{A }+ P_{B} )

=180 **/ **(180+420)

= 180/600

= 0.30

And, mole fraction of liquid B = 1 - 0.30

= 0.70

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Bhavesh
2019-10-14 10:34:20

Aneesha ,what you have a doubt is another representation of the same formulað. p(total)=p(a)+p(b) which is pÂ°(a)x(a)+pÂ°(b)x(b) But we also know that sum of mole fractions is 1 x(b)=1-x(1) Deducing it we get the simpler form

Lia
2019-06-18 19:06:58

Similar doubt as that of anisha's

Tayaba
2019-05-13 13:58:30

What does it mean by composition ? Either it mean mole fraction or vapour pressure ?

rahul
2019-04-24 21:58:38

the query of aneesha is a poor one... Ask something good baby...

Shivam
2018-06-12 20:52:10

It ask for composition. In your answer you give partial vp so it means composition means partial vapour pressure.

Aneesha
2018-03-25 15:15:00

Hello. I have a small query. By using the formula P(total)=P(A)+[P(B)-P(A)]X(B), we get the value of X(B) as 0.42 and X(a) as 0.58. If done in this way, we end up multiplying 450 with 0.58 and 700 with 0.42. Why then am i ending up in the reverse order ? (the formula is given in the NCERT textbook-in terms of 1 and 2 instead of A and B)

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