JEE MAIN - Chemistry (2025 - 7th April Evening Shift - No. 13)

Match List - I with List - II.

List - I List - II
(A) Solution of chloroform and acetone (I) Minimum boiling azeotrope
(B) Solution of ethanol and water (II) Dimerizes
(C) Solution of benzene and toluene (III) Maximum boiling azeotrope
(D) Solution of acetic acid in benzene (IV) ΔVmix = 0

Choose the correct answer from the options given below :

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

Explanation

To correctly match the items from List - I with those in List - II, let's analyze the characteristics of each solution:

(A) Solution of chloroform and acetone: This solution exhibits negative deviation from Raoult's law. It creates a maximum boiling azeotrope because the interactions between chloroform and acetone molecules are stronger than those in the pure components.

(B) Solution of ethanol and water: This solution shows positive deviation from Raoult's law, leading to the formation of a minimum boiling azeotrope. The interactions between ethanol and water molecules are weaker than those in their pure states.

(C) Solution of benzene and toluene: This combination forms an ideal solution where Raoult’s law is obeyed across all concentrations. Thus, the volume change upon mixing, denoted as $\Delta V_{\text{mix}}$, is zero.

(D) Solution of acetic acid in benzene: In this mixture, acetic acid tends to dimerize. The acetic acid molecules pair up, forming dimers, especially in non-polar solvents like benzene.

Based on these explanations, the correct matches are:

(A) - (III)

(B) - (I)

(C) - (IV)

(D) - (II)

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