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Raoult’s Law for a Solution of Two Misci...

Raoult’s Law for a Solution of Two Miscible Liquid || Vapour Pressure and Liquid Composition Curve || Calculation of Mole Fraction in Vapour Phase

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The liquids X and Y from ideal solution having vapour pressures 200 and 100 mm Hg respectively. Calculate the mole fraction of component X in vapour phase in equilibrium with an equimolar solution of the two .

The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :

At 298K , the vapour pressure of pure liquid n-butane is 1823 torr and vapour pressure of pure n-pentane is 521 torr and form nearly an ideal solution. a. Find the total vapour pressure at 298 K of a liquid solution containing 10% n-butane and 90% n-pentane by weight, b. Find the mole fraction of n-butane in solution exerting a total vapour pressure of 760 torr. c. What is composition of vapours of two components (mole fraction in vapour state)?

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A mixture of two miscible liquids A and B is distilled under equilibrium conditions at 1 atm pressure. The mole fraction of A in solution and vapour phase are 0.30 and 0.60 respectively. Assuming ideal behaviour of the solution and the vapour, calculate the ratio of the vapour pressure of pure A to that of pure B.

Vapour Pressure|Factors Affecting Vapour Pressure|Vapour Pressure Of Solutions (Raoult's Law)|Questions|Relation Between Mole Fraction In Solution Phase And Vapour Phase|Summary