Question

Two Liquids with pure state vapour pressure in ratio 2:1 with molar ratio 1:2 moles in vapour pressure

Answer 1

Vapour pressure of liquid 1 , P1 = 2 units

Vapour pressure of liquid 2, P2= 1 units

Moles of liquid 1 = 1

Moles of liquid 2 = 2

mole fraction of liquid 1 = 1 / 3

mole fraction of liquid 2 = 2 / 3

Using Raoult' s law :

Partial pressure of liquid 1 = Vapour pressure of liquid 1

Partial pressure of liquid 1 = 2 ( 1/ 3 ) = 2 / 3

Partial pressure of liquid 2 = Vapour pressure of liquid 2 mole fraction of liquid 2

Partial pressure of liquid 2 = 1 ( 2 / 3 ) = 2 / 3

Total pressure p = partial pressure of 1 + partial pressure of 2 = 2/3+2/3=4/3

Mole fraction of liquid 1 in vapour phase = partial pressure of liquid 1 / total vapour pressure = (2/3) / (4/3) = 1/2

Mole fraction of liquid 2 in vapour phase = partial pressure of liquid 2 / total vapour pressure = (2/3) / (4/3) = 1/2

•°•Ratio of moles = 1 : 1


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