The vapour pressure of a dilute aqueous solution of glucose is 750 mm of Hg at 373 K. The mole fraction of solute in the solution is-
1. 1/10
2. 1/76
3. 1/7.6
4. 1/35

The ratio of the vapour pressure of a solution to the vapour pressure of the solvent is

1. Equal to the mole fraction of the solvent

2. Proportional to the mole fraction of the solute

3. Equal to the mole fraction of the solute

4. None of the above

Mole fraction of the component A in vapour phase is X₁ and mole fraction of component A in liquid mixture is X₂ then 

(${\mathrm{P}}_{\mathrm{A}}^0$ = vapour pressure of pure A; ${\mathrm{P}}_{\mathrm{B}}^0$= vapour pressure of pure B),

then total vapour pressure of the liquid mixture is-

1. $\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_2}{{\mathrm{X}}_1}$

2. $\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_1}{{\mathrm{X}}_2}$

3. $\frac{{\mathrm{P}}_{\mathrm{B}}^0{\mathrm{X}}_1}{{\mathrm{X}}_2}$

4. $\frac{{\mathrm{P}}_{\mathrm{B}}^0{\mathrm{X}}_2}{{\mathrm{X}}_1}$

The vapour pressure of pure water at 298K is 23.76mm. The vapour pressure of a solution of sucrose(C₁₂H₂₂O₁₁) in 5.56 moles of water is 23.392mm Hg. The mass of sucrose in the solution is

1. 15 g

2. 45 g

3. 30 g

4. 37.5 g

The vapour pressure of a pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm.
Mole fraction of the component B in the solution is:

The vapour pressure of water depends upon:

(1) surface area of container

(2) volume of container

(3) Temperature

(4) All of these

The mole(s) of K₂SO₄ to be dissolved in 12 moles of water to lower its vapor pressure by 10 mm of Hg at a temperature at which vapor pressure of pure water is 50 mm of Hg is :

The relative lowering of vapour caused by dissolving 71.3 g of a substance in 1000 g of water is 7.13 x 10^-3. The molecular mass of the substance is-

1. 180 g mol^-1

2. 18 g mol^-1

3. 1.8 g mol^-1

4. 360 g mol^-1

Total vapour pressure of a mixture of 1 mol A (p_A°  = 150 torr) and 2 mol B (${\mathrm{p}}_{\mathrm{B}}^{°}$ = 240 torr) is 200 torr. In this case:-

1. There is a positive deviation from Raoult's law
2. There is a negative deviation from Raoult's law
3. There is no deviation from Raoult's law
4. None of these 

P₁ and P₂ are vapour pressure of solvent and solution at temperature T. The solution is prepared by dissolving non-volatile solute in the same solvent. P₁ =760 torr, P₂ =684 mm Hg, Mole fraction of solvent in the solution will be:-

1. 0.1

2. 0.2

3. 0.9

4. Can't be predicted