What deviation is shown by a mixture of equimolar phenol and aniline?a)Negative deviationb)Positive deviationc)No deviationd)Alternating positive and negativeCorrect answer is option 'A'. Can you explain this answer?

Question

Answer

Explanation:
When two liquids are mixed, the resulting mixture may show either positive or negative deviation from Raoult's law.

- Positive deviation: When the observed vapor pressure of the mixture is higher than the vapor pressure predicted by Raoult's law, the mixture shows positive deviation. This occurs when the intermolecular forces between unlike molecules are weaker than the forces between like molecules, resulting in a higher vapor pressure than expected.
- Negative deviation: When the observed vapor pressure of the mixture is lower than the vapor pressure predicted by Raoult's law, the mixture shows negative deviation. This occurs when the intermolecular forces between unlike molecules are stronger than the forces between like molecules, resulting in a lower vapor pressure than expected.
- No deviation: If the observed vapor pressure of the mixture is equal to the vapor pressure predicted by Raoult's law, the mixture shows no deviation.

In the case of equimolar phenol and aniline, the two molecules have similar molecular structures and can form hydrogen bonds with each other. The hydrogen bonding between phenol and aniline is stronger than the hydrogen bonding within each pure component, resulting in a lower vapor pressure than expected. Therefore, the mixture of equimolar phenol and aniline shows negative deviation from Raoult's law.

Answer

Phenol and aniline both exhibit greater magnitude of hydrogen bonding due to the presence of hydroxyl and amine group, both possessing highly electronegative atoms of oxygen and nitrogen, respectively. The intermolecular hydrogen bonding is far greater than the degree of intramolecular hydrogen bonding which requires more thermal energy to break. Thus, there is an increase in the boiling point of the mixture.

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