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 largest freezing point depression among the following 0.10 m solutions is shown by:

The equal weight of a solute is dissolved in an equal weight of two solvents A and B to form a very dilute solution. The relative lowering of vapour pressure for solution B has twice the relative lowering of vapour pressure for solution A.

If ${\mathrm{M}}_{\mathrm{A}}$ and ${\mathrm{M}}_{\mathrm{B}}$ are the molecular weights of solvents A and B respectively, then:

1. ${\mathrm{M}}_{\mathrm{A}}$ = ${\mathrm{M}}_{\mathrm{B}}$

2. ${\mathrm{M}}_{\mathrm{B}}$ = 2${\mathrm{M}}_{\mathrm{A}}$

3. ${\mathrm{M}}_{\mathrm{A}}$ = 4${\mathrm{M}}_{\mathrm{B}}$

4. ${\mathrm{M}}_{\mathrm{A}}$ = 2${\mathrm{M}}_{\mathrm{B}}$

One mole of sugar is dissolved in three moles of water at 298 K. The relative lowering of vapour pressure is:

Lowering in vapour pressure is the highest for:

1. 0.2 m urea

2. 0.1 m glucose

3. 0.1 m MgSO₄

4. 0.1 m BaCl₂

If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapor pressure to 80 %, the molar mass (in g mol^–1) of the solute is:

[Molar mass of n-octane is 114 g mol^–1]

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

At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be:
(molar mass of urea = 60 g mol^–1)