Vapour pressure of pure liquid A & B are 21 & 18 mm of Hg respectively. Determine vapour pressure of a solution obeying Raoult's law containing 1 mole of A& 2mole of B

At 40^(@)C the vapour pressure of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg respectively. At the same temperature, the vapour pressure of an equimolar solution of the liquids, assuming the ideal solution will be:

The vapour pressure of pure liquid A and liquid B at 350 K are 440 mm and 720 mm of Hg. If total vapour [ressure of solution is 580 mm of Hg then the mole fraction of liquid A in vapour phase will be :-

At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mm Hg, respectively. If in a mixture of a and B, the vapoure is 300 mm Hg, the mole fractions of a in the vapour phase, respectively, are -

The vapour pressure of two pure liquids A and B are 50 and 40 torr respectively. If 8 moles of A is mixed with x moles of B, then vapour pressure of solution obtained is 48 torr. Calculate the value of x.

The vapour pressure of two pure isomeric liquids X and Y are 200 torr and 100 torr respectively at a given temperature. Assuming a solution of these components to obey Raoults law, the mole fraction of component X in vapour phase in equilibrium with the solution containing equal amounts of X and Y, at the same temperature is :