Which of the following liquid pairs shows a positive deviation from Raoult’ law?

Benzene–methanol

The correct answer is option 3. Benzene–methanol.

Let us delve into the reasoning behind positive and negative deviations from Raoult's law and analyze why benzene–methanol exhibits positive deviation.

Raoult's law states that the partial vapor pressure of each component in an ideal solution is directly proportional to its mole fraction. For a binary mixture of components A and B:

\(P_A = P_A^\circ x_A \)

\( P_B = P_B^\circ x_B \)

where \( P_A \) and \( P_B \) are the partial vapor pressures of A and B, \( P_A^\circ \) and \( P_B^\circ \) are the vapor pressures of pure A and B, and \( x_A \) and \( x_B \) are the mole fractions of A and B.

Positive Deviation from Raoult's Law:

Positive deviation occurs when the interactions between unlike molecules (A-B) are weaker than those between like molecules (A-A and B-B). This leads to an increase in the total vapor pressure of the solution, as molecules escape more easily into the vapor phase.

Analysis of Given Pairs:

1. Water–Nitric Acid:

Interactions: Strong hydrogen bonding and ionic interactions between water and nitric acid.

Deviation: Likely to show negative deviation due to stronger interactions between water and nitric acid molecules compared to pure components.

2. Water–Hydrochloric Acid:

Interactions: Strong hydrogen bonding and ionic interactions between water and hydrochloric acid.

Deviation: Likely to show negative deviation due to stronger interactions between water and hydrochloric acid molecules compared to pure components.

3. Benzene–Methanol:

Interactions: Benzene is nonpolar with weak Van der Waals forces, while methanol is polar and can form hydrogen bonds. The interactions between benzene and methanol are primarily Van der Waals forces, which are weaker than the hydrogen bonds between methanol molecules.

Deviation: Likely to show positive deviation because the interactions between benzene and methanol molecules are weaker than the interactions within pure methanol (hydrogen bonding) and pure benzene (Van der Waals forces).

4. Acetone–Chloroform:

Interactions: Strong dipole-dipole interactions and hydrogen bonding. Acetone has a carbonyl group, and chloroform can form hydrogen bonds with it.

Deviation: Likely to show negative deviation due to strong interactions between acetone and chloroform molecules compared to pure components.

Conclusion:

Benzene and methanol show a positive deviation from Raoult's law because the interaction between benzene and methanol molecules (primarily Van der Waals forces) is weaker than the hydrogen bonding present in pure methanol and the Van der Waals forces in pure benzene. This weaker interaction results in an increased tendency for the molecules to escape into the vapor phase, thereby increasing the total vapor pressure above the ideal value predicted by Raoult's law.