The V.P.of water at R.T.is `30mmHg`.In a solution of a non-volatile solute in water,the mole fraction of water is `0.9`.The V.P.of the solution is `(P_(s)=P_(0)X_( solvent )` 1) `3mmHg,` 2) `30mmHg` 3) `21mmHg,` 4) `27mmHg`

2.82g of glucose is dissolved in 30g of water. The mole fraction of glucose in the solution is

Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9 . The boiling point of this solution is

Is p = 2 a solution for 2p-1=3 ?

V.P. of solvent in pure state is 600 mm Hg, when a non-volatile solute is added to it vapour pressure of solution becomes 594 mm Hg, then x_(B) will be

The P^(H) of a solution is 3.0. This solution is diluted by 100 times. Then the P^(H) of the resulting solution is

1 mole of glucose is dissolved in 4 moles of water to form a solution. Find mole-fraction of solute in solution (x_("solute")) ?

The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of solute is 0.2 , what would be the mole fraction of solvent if the decrease in vapour pressure is 20 mm of Hg .

The P^(H) of a I lit solution is 2. It is diluted with water till its p^(H) becomes 4, How many litres of water is added?

Assuming ideal behaviour, calculate the pressure of 1.0 molal solution of a non-volatile molecular solute in water at 50@C . The vapour presure of water at 50@C is 0.222 atm .

Mole fraction of vapour of A above solution in mixture of A and B(X_(A) = 0.4) will be (P_(A)^(@) = 100mm, P_(B)^(@) = 200 mm) :