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

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:

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

An aqueous solution is 1.00 molal in KI. The vapour pressure of the solution can be increased by:

1. Addition of NaCl

2. Addition of Na₂SO₄

3. Addition of 1.00 molal Kl

4. Addition of water

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${\mathrm{M}}_{\mathrm{B}}$