NCERT Class XII Chemistry
Chapter - Solutions
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Question : 49
Total: 53
The vapour pressures of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Solution:
Given : P 0 A = 450 m m H g ,
P 0 B = 700 m m H g ,
P Total = 600 m m H g , x A = ?
Applying Raoult's law,P A = x A × P 0 A
P B = x B × P 0 B = ( 1 − x A ) P 0 B
P T o t a l = P A + P B = x A × P 0 A + ( 1 − x A ) P 0 B
= P 0 B + P 0 A − P 0 B ) x A
Substituting the given values, we get
600 = 700 + ( 450 − 700 ) x A
or,250 x A = 100
orx A =
= 0.40
Thus, composition of the liquid mixture will be
x A ( A ) = 0.40
x B ( B ) = 1 − 0.40 = 0.60
Calculation of composition in the vapour phase
P A = x A × P 0 A = 0.40 × 450 m m H g = 180 m m H g
P B = x B × P 0 B = 0.60 × 700 m m H g = 420 m m H g
Mole fraction ofA =
=
= 0.30
Mole fraction of B in the vapour phase= 1 – 0.30 = 0.70
Applying Raoult's law,
Substituting the given values, we get
or,
or
Thus, composition of the liquid mixture will be
Calculation of composition in the vapour phase
Mole fraction of
Mole fraction of B in the vapour phase
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