An ideal solution was found to have a vapour pressure of 80 torr when the mole fraction of a non-volatile solute was 0.2. What would be the vapour pressure of the pure solvent at the same temperature?

If to a solvent (vapour pressure =p^(@)) , n moles of a non volatile solute are dissolved then the vapour pressure of the solution p is

Relative loweringof vapour pressure of a dilute solution is 0.2 . What is the mole fraction of non-volatile solute ?

Relative loweringof vapour pressure of a dilute solution is 0.2 . What is the mole fraction of non-volatile solute ?

The vapour pressure at a given temperature of an ideal solution containing 0.2 mol of non-volatile solute and 0.8 mol of a solvent is 60 mm of Hg . The vapour pressure of the pure solvent at the same temperature will be

The vapour pressure at a given temperature of an ideal solution containing 0.2 mol of non-volatile solute and 0.8 mol of a solvent is 60 mm of Hg . The vapour pressure of the pure solvent at the same temperature will be

The vapour pressure at a given temperature of an ideal solution containing 0.2 mol of non-volatile solute and 0.8 mol of a solvent is 60 mm of Hg . The vapour pressure of the pure solvent at the same temperature will be

At room temperature the mole fraction of a solution is 0.25 and the vapour pressure of the solvent is 0.80 atm. Then the lowering of vapour pressure is

The vapour pressure of a solution containing non volatile solute is less than the vapour pressure of pure of solvent. Give reason.