What type of deviation from Raoult's law is shown by a mixture of phenol and aniline? Give reason.
What will happen to the boiling point of the solution on mixing phenol and aniline?
What type of deviation from Raoult's law is shown by a mixture of phenol and aniline? Give reason.
What will happen to the boiling point of the solution on mixing phenol and aniline?
Official answer
Lead: A phenol–aniline mixture shows negative deviation from Raoult's law.
Reason:
- The –OH group of phenol and the –NH₂ group of aniline form new intermolecular hydrogen bonds (O–H···N).
- These A–B attractions are stronger than the A–A and B–B attractions in the pure liquids.
- The escaping tendency of the molecules falls, so the vapour pressure is lower than predicted by Raoult's law → negative deviation.
Effect on boiling point:
- Because the vapour pressure is lowered, the boiling point increases (rises above that of the individual components).
Marking scheme
- 1.1 mark: identifying negative deviation from Raoult's law with correct reason (strong intermolecular hydrogen bonding between phenol –OH and aniline –NH₂ lowers vapour pressure).
- 2.1 mark: stating that the boiling point of the solution increases (rises) because vapour pressure is lowered.
- 3.Accept equivalent wording: 'intermolecular H-bonding stronger than in pure components' and 'boiling point higher than either component'.
Hint
Ask whether new A–B forces (here O–H···N hydrogen bonds) are stronger than the forces in the pure liquids — stronger means negative deviation and a higher boiling point.
Quick oral answer
A phenol–aniline mixture shows negative deviation from Raoult's law because strong O–H···N hydrogen bonding between the –OH of phenol and the –NH₂ of aniline lowers the vapour pressure, so the boiling point of the solution increases.
Common mistakes
- 1.Calling it positive deviation — the new O–H···N hydrogen bonds are stronger than the interactions in the pure liquids, giving negative deviation.
- 2.Saying the boiling point decreases; lowered vapour pressure means the boiling point rises.
- 3.Omitting the reason (hydrogen bonding) and stating only the type of deviation, losing half the marks.
Analysis and explanation
Concept:
Deviations from Raoult's law depend on the relative strengths of solute–solvent (A–B) versus pure-component (A–A, B–B) interactions.
Why negative deviation here:
- Phenol supplies an acidic –OH hydrogen; aniline supplies a basic nitrogen lone pair on –NH₂.
- Together they form strong intermolecular hydrogen bonds that did not exist (or were weaker) in the pure liquids.
- Stronger A–B forces reduce escaping tendency → observed vapour pressure < Raoult's-law prediction.
Consequences of negative deviation:
- Vapour pressure is lower than expected.
- Boiling point is higher than expected.
- Mixing is exothermic () and volume contracts slightly ().
Exam trap:
Students sometimes label it positive deviation by analogy with ethanol–water. Here the NEW O–H···N bonding is stronger than the interactions in the separate liquids, so it is firmly a negative deviation.
Real-world:
Strong negative-deviation pairs (like nitric acid–water) form maximum-boiling azeotropes, which is why such mixtures boil at a temperature higher than either pure component.
Interesting facts
Phenol–aniline is a textbook example of a negative-deviation pair precisely because an acidic O–H and a basic N combine to hydrogen bond strongly.
Negative-deviation mixtures release heat on mixing (exothermic) and contract in volume, the opposite of positive-deviation mixtures like ethanol–acetone.
Question details
Q17 of 37 questions
- Board
- CBSE
- Class
- Class 12
- Subject
- Chemistry
- Year
- 2026
- Section
- Section B
- Marks
- 2
- Type
- Very Short Answer
- Chapter
- Solutions
- Topic
- Non-ideal solutions: negative deviation
- Bloom's level
- understand
- Syllabus topic
- Unit: Solutions — Non-ideal Solutions and Deviations from Raoult's Law
- Author
- CBSE
- Reviewed by
- Padhantu Academic Team
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Frequently asked questions
Why does a phenol–aniline mixture show negative deviation from Raoult's law?
Because the acidic –OH of phenol and the basic –NH₂ of aniline form strong intermolecular O–H···N hydrogen bonds. These A–B attractions are stronger than the attractions in the pure liquids, so the escaping tendency and vapour pressure drop below the Raoult's-law prediction.
Does the boiling point rise or fall for the phenol–aniline mixture?
It rises. Negative deviation means the vapour pressure is lower than expected, and a lower vapour pressure corresponds to a higher boiling point — often higher than that of either pure component.