Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(II), (B)-(I), (C)-(III), (D)-(IV) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
(A)-(IV), (B)-(III), (C)-(II), (D)-(I) |
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(A)-(IV) Pure ethanol forms strong $\text{H}$-bonds. When water is added, the original $\text{H}$-bonds in ethanol are broken, and new $\text{H}$-bonds between ethanol and water are weaker than the original ones. This leads to large positive deviation. (B)-(III) In the mixture, new, strong intermolecular $\text{H}$-bonds form between the $\text{H}$ of chloroform ($\text{CHCl}_3$) and the $\text{O}$ of acetone ($\text{CH}_3\text{COCH}_3$). This increased attraction lowers the escaping tendency of molecules, causing a negative deviation. (C)-(II)These two are non-polar organic liquids with very similar molecular size and weak Van der Waals forces. The $\text{A-A}$ and $\text{B-B}$ interactions are nearly identical to the $\text{A-B}$ interactions. This forms an ideal solution with no deviation. (D)-(I)Both components have weak intermolecular forces. When mixed, the new $\text{A-B}$ interactions are weaker than the strong dipole-dipole interactions in pure acetone. This makes molecules escape more easily, resulting in a positive deviation. |