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 :
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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