Question

Which one of the following binary liquid mixtures exhibit positive deviation from Raoult’s law?

• A. Carbon disulphide – acetone
• B. Chloroform – acetone
• C. Bromobenzene – chlorobenzene
• D. Benzene – toluene
• E. Phenol – aniline

Hint:

Weak A–B forces → positive deviation; strong A–B forces (like hydrogen bonding) → negative deviation.

Answer 1

The Correct Option is A

Concept: Raoult’s law states that the partial vapour pressure of a component in an ideal solution is proportional to its mole fraction: \[ P_i = X_i P_i^0 \] However, real solutions deviate from this law due to differences in intermolecular forces. Positive deviation occurs when:
• The intermolecular attraction between unlike molecules (A–B) is weaker than that between like molecules (A–A and B–B).
• As a result, molecules escape more easily into vapour phase.
• Vapour pressure becomes greater than expected. Negative deviation occurs when:
• A–B interactions are stronger (e.g., hydrogen bonding).
• Vapour pressure becomes lower than ideal. Step-by-step analysis of each option: (A) Carbon disulphide – acetone:
• CS$_2$ is non-polar, acetone is polar.
• Interactions between them are weaker than their own interactions.
• This weak attraction allows molecules to escape easily.
• Hence, positive deviation. (B) Chloroform – acetone:
• Strong hydrogen bonding occurs between CHCl$_3$ and acetone.
• This increases intermolecular attraction.
• Vapour pressure decreases → negative deviation. (C) Bromobenzene – chlorobenzene:
• Similar size and polarity.
• Nearly ideal solution.
• No significant deviation. (D) Benzene – toluene:
• Both are non-polar hydrocarbons.
• Interactions are almost identical.
• Ideal solution. (E) Phenol – aniline:
• Strong hydrogen bonding between OH and NH$_2$ groups.
• Strong attraction reduces escaping tendency.
• Negative deviation. Final Conclusion: Only option (A) shows weaker A–B interactions, hence: \[ \boxed{\text{Carbon disulphide – acetone}} \]