JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
Match items of column I and II
Correct match is
| Column I (Mixture of compounds) | Column II (Separation Technique) | ||
|---|---|---|---|
| A. | H₂O/CH₂Cl₂ | i. | Crystalization |
| B. | ii. | Differential solvent extraction | |
| C. | Kerosene / Naphthalene | iii. | Column chromatography |
| D. | C₆H₁₂O₆/NaCl | iv. | Fractional Distillation |
- (a) A-(ii), B-(iv), C-(i), D-(iii)
- (b) A-(iii), B-(iv), C-(ii), D-(i)
- (c) A-(i), B-(iii), C-(ii), D-(iv)
- (d) A-(ii), B-(iii), C-(iv), D-(i)
Answer and solution
Answer (d) A-(ii), B-(iii), C-(iv), D-(i)
Solution
Chemistry ma’am · 3 steps
Ask what each pair is physically: two liquids, two solids, or two compounds of different polarity.
- Match by property. Every separation technique uses one difference between the two components. Find that difference for each mixture, and the technique follows.
Mixture What differs Technique A. H₂O/CH₂Cl₂ Two immiscible liquids that form separate layers (ii) differential solvent extraction B. α-tetralone / p-nitrophenol Different polarity (ketone versus phenol with -NO₂), so they adsorb differently on silica (iii) column chromatography C. Kerosene / naphthalene Naphthalene dissolves in kerosene. Both are volatile, with boiling points that are not far apart (iv) fractional distillation D. C₆H₁₂O₆/NaCl Two non-volatile solids with different solubility in a chosen solvent (i) crystallisation - Rule out quickly. A must be (ii), which leaves only (a) and (d). C must be (iv), since kerosene is a liquid and cannot be crystallised. That removes (a).
- So. The correct match is A-(ii), B-(iii), C-(iv), D-(i), which is option (d).
The idea to keep. Immiscible liquids: solvent extraction. Miscible liquids with close boiling points: fractional distillation. Solids differing in solubility: crystallisation. Different polarity: chromatography.
Answer from the NTA final answer key. Solution by PrepWiser.