Question

Ideal mixture of two miscible liquids A and B is placed in a cylinder containing piston. Piston is pulled out isothermally so that volume of liquid decreases but that of vapours increases. When negligibly small amount of liquid was left, the mole fraction of A in vapour phase is 0.4. If P°a = 0.4 atm and P% = \2 atm at the experimental temperature, calculate the total pressure at which the liquid is almost evaporated.

Answer 1

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Two liquids A and B on mixing form ideal solution. their vapour pressures in pure state are 200 and 100 mm respectively. what will be the mole frction of B in the vapour phase in equillibrium with an equimolar solution of two?

Ask for details Follow Report by Akshay233 15.05.2018

Answers

Aarshika Ambitious

Let the moles of A=B = m

Mole fraction of B = m/m+m =m/2m

According to Raoults law

P=P°x

P(B)=P°(B)x(B)

x(B) = m/2m

P°(B) = 100

P(B) = 100*m/2m = 50mm

Similarly

P(A) = 100mm

Y(B) = P(B)/P(A)+P(B)

= 50/100+50 = 1/3

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