CUET Preparation Today
Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(I), (B)-(II), (C)-(III), (D)-(IV) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
(A)-(III), (B)-(II), (C)-(I), (D)-(IV) |
The correct answer is Option (3) → (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
Solutions behave differently based on intermolecular interactions. • Strong interactions → maximum boiling azeotrope • Weak interactions → minimum boiling azeotrope • Similar molecules → ideal solution (A) Phenol and Aniline: These two molecules form strong intermolecular hydrogen bonds with each other. The oxygen of the hydroxyl group in phenol interacts with the hydrogen of the amino group in aniline. This strong interaction leads to a negative deviation from Raoult's Law.
(B) Nitric acid and water: This mixture shows a large negative deviation from Raoult's Law because the interactions between $HNO_3$ and $H_2O$ are stronger than the pure components. Such solutions form a maximum boiling azeotrope, which boils at a temperature higher than either pure component.
(C) Ethanol and water: This mixture exhibits a large positive deviation from Raoult's Law because the ethanol-water interactions are weaker than the pure liquid interactions. This results in a minimum boiling azeotrope, which boils at a temperature lower than either ethanol or water.
(D) Benzene and toluene: These are both non-polar hydrocarbons with very similar structures and molecular sizes. Because the $A-B$ interactions are almost identical to the $A-A$ and $B-B$ interactions, they form a nearly ideal solution.
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