At 300 K two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg. In a equimolar liquid mixture of A and B, the mole fraction of B in the vapour phase above the solution at this temperature is:

At 300 K, two pure liquids A and B have vapour pressures 150 mm Hg and 100 mm Hg respectively. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is:

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 -

At a particular temperature, the vapour pressure of two liquids A and B are 120 and 180 mm Hg respectively. If 2 mole of A and 3 mole of B are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in mm Hg)

Two liquids A and B with vapour pressures 400mm Hg and 600mm Hg respectively formed a binary ideal liquid mixure shown in the following diagram . (Temepature =constant) If the mole fraction of 'A' in the vapour phase is 0.5 , What is the optimum pressure , above which the vapour phase can start condensig back into liquid.

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

The vapour pressure of pure benzene and toluene are 160 and 60 mm Hg respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is :

The vapour presssures of pure benzene and toluene are 160 and 60 mm Hg respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is