Q. 84.4( 59 Votes )

# 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 :

Given,

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

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

p_{total} = 600 mm of Hg

By using Rault's law,

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

p_{total} = P_{A}^{o}x_{A} + P_{B}^{o}x_{B}

p_{total} = P_{A}^{o}x_{A} + P_{B}^{o}( 1 - x_{A} )

p_{total} = (P_{A}^{o}- P_{B}^{o})x_{A} + P_{B}^{o}

600 = (450 - 700) x_{A} + 700

-100 = -250 x_{A}

x_{A} = 0.4

∴ x_{B} = 1 - x_{A}

x_{B} = 1 – 0.4

x_{B} = 0.6

Now,

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

P_{A} = 450 × 0.4

P_{A} = 180 mm of Hg

and

P_{B} = P_{B}^{o}x

P_{B} = 700 × 0.6

P_{B} = 420 mm of Hg

Composition in vapour phase is calculated by

Mole fraction of liquid,

=

= 0.30

Mole fraction of liquid,

=

= 0.70

__Note:__ Alternate method to find the Mole fraction of liquid B is

= 1 - Mole fraction of liquid A

= 1 – 0.30

= 0.70

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