An ideal solution is one in which the attraction between components of the solution is the same as the interaction between the molecules of each component. Heat is neither absorbed nor evolved during the formation of an ideal solution and the volume of the solution is equal to the sum of the volumes of the component liquids. Raoult’s law states that the partial vapour pressure of each component (say liquid A) of the solution is directly proportional to its mole fraction in the solution. The total vapour pressure over the solution is equal to the sum of the partial vapour pressures of the individual components.
A non-ideal solution is that solution (i) which does not obey Raoult’s law (ii) for which \(\Delta V_{mix}\) is not zero and (iii) for which \(\Delta H_{mix}\) is not zero. In non-ideal solutions, the solute–solvent interactions are weaker or stronger than the solute–solute and solvent–solvent interactions. The non-ideal solutions in which solute–solvent interactions are weaker than the solute–solute and solvent–solvent interactions show positive deviation from Raoult’s law, and those in which they are stronger show negative deviation.
If two liquids A and B form a minimum boiling azeotrope at a specific composition, then:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A–B interactions are weaker than those between A–A or B–B.
A minimum boiling azeotrope is formed by a solution that shows a large positive deviation from Raoult’s law.
In positive deviation the A–B attractions are weaker than the A–A and B–B attractions. Molecules of A and of B escape more easily than from the pure liquids, so the vapour pressure is higher than Raoult’s law predicts. A higher vapour pressure reaches the outside pressure at a lower temperature, so at one composition the mixture boils below both pure liquids.
Why the other options are incorrect
Option (1) A–B interactions are stronger than those between A–A or B–B. — Incorrect. Stronger A–B attractions lower the escaping tendency and give negative deviation, which leads to a maximum boiling azeotrope.
Option (2) The solution shows a negative deviation from Raoult’s law. — Incorrect. A large negative deviation lowers the vapour pressure and forms a maximum boiling azeotrope, not a minimum boiling one.
Option (3) The vapour pressure of the solution decreases because fewer molecules of only one of the liquids escape from the solution. — Incorrect. In a minimum boiling azeotrope the vapour pressure increases, and molecules of both A and B escape more easily.
NCERT (2026-27 edition):
1.5.2 Non-ideal Solutions – “… In case of positive deviation from Raoult’s law, A-B interactions are weaker than those between A-A or B-B, i.e., in this case the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. … The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a specific composition.”