Positive deviation from Raoult's law are exhibited by binary liquid mixtures

Options

A. in which the molecules tend to attract each other and hence their escape into the vapor phase is retarted.
B. in which the molecules repel each other and hence enter the vapor phase more readly than do the molecules of the pure liquids.
C. In which the molecules attract each other and hence enter the vapour phase more readly than do the molecules of the pure liquids.
D. In which the molecules repel each other and hence do not enter the vapor phase as the molecules of the pure liquids do.

Answer

2

Explanation

If the intermolecular force between `A` and `B` molecules are weaker than those between `A` molecules and between `B` molecules, then there is a greater tendency for these molecules to leave the solution than in the case of an ideal solution. Consequently, the vapor pressures of the solution is greater than the sum of the vapor pressure as predicted by Raoult's law for the same concentration. This behavior gives rise to the positive deviation.
Thus, a solution of a pair of volatile liquids `A` and `B` will show positive deviation from Raoult's law if
(a) `P_(A) gt P_(A)^(0) chi_(A)` and `P_(B) gt P_(B)^(0) chi_(B)`
(b) The intermolecular forces of `A-A, B-B gt A-B`
(c ) `Delta_(mix)H` is `+ve, Delta_(mix)V is +ve`
If `A` molecules attract `B` molecules more strongly than they do their own kind, the vapor pressure of the solution is less than the sum of the vapor pressure as predicted by Raoult's law. Here we have a negative deviation.
Thus, a solution of a pair of volatile liquid `A` and `B` shows negative deviation from Raoult's law if
(a) `P_(A)lt P_(A)^(0) chi_(A)` and `P_(B)lt P_(B)^(0) chi_(B)`
(b) The intermolecular forces `A-A, B-B lt A-B`.
(c ) Both `Delta_(mix)H` and `Delta_(mix) V` are negative