What will be the relative lowering in vapour pressure when 1 mole of A(l) is dissolved in 9 moles of B(l). Given both are volatile with vapour pressures as 100 mm of Hg and 300 mm of Hg respectively?

Two liquids A and B form an ideal solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively. Then x is-

Two liquids A and B form an ideal solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively. Then x is-

The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is

The vapour pressure of a solution contain2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is

What will be the number of moles of K_(2)SO_(4) to be dissolved in 12 mol of water to lower . Its vapour pressure by 10 mm Hg at a temperature at which vapours pressure of pure water is 50 mm .

The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of the water is 0.9, the vapour pressure of the solution will be :

At 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure of the solution increases by 25 mm of Hg, if one more mole of B is added to the above ideal solution at 300K. Then the vapour pressure of A in its pure state is

A 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure of the solution increases by 25 mm of Hg if one more mole of B is added to the above ideal solution at 300 K. Then vapour pressure of A in its pure state is :