An ideal solution of A and B contains 33.33 mole percent of A in liquid phase and is in equilibrium with its vapour that contains 33.33 mole percent of B .Calculate the ratio of vapour pressure of pure A and B .

When an ideal binary solution is in equilibrium with its vapour, molar ratio of the two components in the solution and in the vapur phase is :

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.

Two liquids A and B are mixed to form an ideal solution. The totol vapour pressure of the solution is 800 mm Hg. Now the mole fractions of liquid A and B are interchanged and the total vapour pressure becomes 600 mm of Hg. Calculate the vapour pressure of A and B in pure form. (Given : p_(A)^(@)-p_(B)^(@)=100 )

A certain ideal solution of two liquids A and B has mole fraction of 0.3 and 0.5 for the vapour phase and liquid phase, respectively. What would be the mole fraction of B in the vapour phase, when the mole fraction of A in the liquid is 0.25 ?

At 25^(@)C , the total pressure of an ideal solution obtained by mixing 3 mole of A and 2 mole of B, is 184 torr. What is the vapour pressure (in torr) of pure B at the same temperature (Vapour pressure of pure A at 25^(@)C is 200 torr) ?

At 27^(@)C .the vapour pressure of an ideal solution containing 1 mole of A and 1 mole and B is 500 mm of Hg . At the same temperature, if 2 mol of B is added to this solution the vapour pressure of solution increases by 50 mm of Hg . The vapour pressure of A and B in their pure states is respectively.

Two solutions of A and B are available. The first is known to contain 1 mole of A and 3 moles of B and its total vapour pressure is 1.0 atm. The second is known to contain 2 moles of A and 2 moles of B , its vapour pressure is greater than 1 atm, but if is found that this total vapour pressure may be reduced to 1 atm by the addition of 6 moles of C . The vapour pressure of pure C is 0.80 atm. Assuming ideal solutions and that all these data refered to 25^(@)C , calculate the vapour pressure of pure A and B .

An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of C, P_(C)^(@)=0.8 atm. Calculate the vapour pressure of pure A and B :

At 300 K , the vapour pressure of an ideal solution containing one mole of A and 3 mole of B is 550 mm of Hg . At the same temperature, if one mole of B is added to this solution, the vapour pressure of solution increases by 10 mm of Hg . Calculate the V.P. of A and B in their pure state.

In a mixture of A and B having vapour pressure of pure A and B are 400 m Hg and 600 mm Hg respectively. Mole fraction of B in liquid phase is 0.5. Calculate total vapour pressure and mole fraction of A and B in vapour phase .