Logo
Jeeify
MCQMediumJEE 2025Raoult's Law & Vapour Pressure

JEE Chemistry 2025 Question with Solution

Match List-I with List-II

Table showing List-I and List-II for matching: chloroform and acetone, ethanol and water, benzene and toluene, acetic acid in benzene with azeotrope and solution properties.

Choose the correct answer from the options given below :

  • A

    (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

  • B

    (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

  • C

    (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

  • D

    (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A matching problem based on properties of solutions and deviations from Raoult's law.

Find: The correct correspondence between List-I and List-II.

For the listed pairs:

  1. Solution of chloroform and acetone shows strong hydrogen bonding between chloroform and acetone. Stronger intermolecular attraction causes negative deviation from Raoult's law, so it forms a maximum boiling azeotrope. Therefore, (A) (\to) (III).

  2. Solution of ethanol and water shows positive deviation from Raoult's law and forms a minimum boiling azeotrope. Therefore, (B) (\to) (I).

  3. Solution of benzene and toluene behaves nearly ideally because the intermolecular forces are similar. For an ideal solution,

ΔVmix=0\Delta V_{\text{mix}} = 0

Therefore, (C) (\to) (IV).

  1. Solution of acetic acid in benzene shows association of acetic acid molecules in a non-polar solvent, so it dimerizes. Therefore, (D) (\to) (II).

Thus, the correct arrangement is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Therefore, the correct option is A.

Worked solution image summarizing the matches between listed solutions and properties like azeotropes, dimerization, and delta V mix equals zero.

Concept-Based Matching

Given: Four solutions are to be matched with their characteristic behaviors.

Find: The correct matching pair set.

Step 1: Analyze chloroform + acetone.

Chloroform and acetone exhibit strong hydrogen bonding. This makes the escaping tendency of molecules smaller than expected, causing negative deviation from Raoult's law. Such solutions form a maximum boiling azeotrope.

So, (A) (\to) (III).

Step 2: Analyze ethanol + water.

This mixture shows positive deviation from Raoult's law and hence forms a minimum boiling azeotrope.

So, (B) (\to) (I).

Step 3: Analyze benzene + toluene.

These are structurally similar non-polar liquids with similar intermolecular forces. Hence, the solution behaves nearly ideally, and for an ideal solution,

ΔVmix=0\Delta V_{\text{mix}} = 0

So, (C) (\to) (IV).

Step 4: Analyze acetic acid in benzene.

In a non-polar solvent like benzene, acetic acid associates through hydrogen bonding and forms dimers.

So, (D) (\to) (II).

Step 5: Write the final matching.

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Therefore, the correct option is A.

Common mistakes

  • Confusing positive deviation with maximum boiling azeotrope is incorrect because positive deviation increases vapour pressure and leads to a minimum boiling azeotrope. Always relate stronger interactions to negative deviation and weaker interactions to positive deviation.

  • Assuming benzene and toluene form a non-ideal solution is incorrect because their intermolecular forces are very similar. Treat this pair as nearly ideal and use ΔVmix=0\Delta V_{\text{mix}} = 0.

  • Missing the dimerization of acetic acid in benzene is a conceptual error because acetic acid associates in non-polar solvents through hydrogen bonding. Do not treat it as existing only as monomer molecules in benzene.

Practice more Raoult's Law & Vapour Pressure questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Raoult's Law & Vapour Pressure

Related questions