Class 12 Chemistry - Chapter Solutions NCERT Solutions | The vapour pressure of pure liquids A an
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.
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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6 Comment(s) on this Question
Aneesha ,what you have a doubt is another representation of the same formulað. p(total)=p(a)+p(b) which is p°(a)x(a)+p°(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
Similar doubt as that of anisha's
What does it mean by composition ? Either it mean mole fraction or vapour pressure ?
the query of aneesha is a poor one... Ask something good baby...
It ask for composition. In your answer you give partial vp so it means composition means partial vapour pressure.
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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