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

The van’t Hoff factor (i) for a dilute aqueous solution of a strong electrolyte barium hydroxide is:

1. 0
2. 1
3. 2
4. 3

Consider the following statements about the composition of the vapour over an ideal 1:1 molar mixture of benzene and toluene. The correct statement is:

Assume that the temperature is constant at 25 ^oC.
(Given, vapour pressure data at 25 °C, benzene = 12.8 kPa, toluene = 3.85 kPa)

The boiling point of 0.2 mol kg^–1 solution of X in water is greater than the equimolal solution of Y in water. The correct statement in this case is:

The electrolyte having the same value of Van't Hoff factor (i) as that of Al₂(SO₄)₃  (if all are 100% ionized) is:

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

The freezing point depression constant for water is 1.86 ^oC m^-1. If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by -3.82 ^oC. The Van’t Hoff factor for Na₂SO₄ is: