Q 2.49 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.
Q 2.49 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.
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1 Answer
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Given, PAo = 450 mm Hg
PBo = 700 mm Hg
ptotal = 600 mm of Hg
By using Rault's law,
ptotal = PA + PB
ptotal = PAoxA + PBoxB
ptotal = PAoxA + PBo ( 1 - xA )
ptotal = (PAo- PBo)xA + PBo
600 = (450 - 700) xA + 700
-100 = -250 xA
xA = 0.4
∴ xB = 1 - xA
xB = 1 – 0.4
xB = 0.6
Now,
PA = PAoxA
PA = 450 × 0.4
PA = 180 mm of Hg and
PB = PBox
PB = 700 × 0.6
PB = 420 mm of Hg
Composition in vapour phase is calculated by
Mole fraction of liquid,
A =PA / PA + PB
= 180/180+420
= 0.30
Mole fraction of liquid,
B =PB / PA + PB
= 420 / 180+420
= 0.70
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