|
List-I (Solution) |
List-II (nature of solution) |
| (P) HNO3 and water | (1) Positive deviation from Raoult's Law |
| (Q) Ethanol and Acetone | (2) Molarity = Molality |
| (R) 5 M NaOH solution having d = 1.25 gm /ml | (3) Actual colligative properties will be different as compare to expected colligative properties. |
| (S) 5 × 10–3 M aqueous KOH solution | (4) Negative deviation from Raoult's Law |
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This question matches solutions (List-I) with their properties (List-II) based on colligative behavior and deviations from Raoult's Law.
(P) HNO3 and water forms a maximum boiling azeotrope due to strong hydrogen bonding, showing (4) Negative deviation from Raoult's Law (vapor pressure lower than ideal).
(Q) Ethanol and Acetone have weaker intermolecular forces than pure components, showing (1) Positive deviation (vapor pressure higher than ideal).
(R) 5 M NaOH (d=1.25 g/ml) is a concentrated electrolyte solution where ions associate, so (3) Actual colligative properties differ from expected values due to van't Hoff factor (i) not equal to number of ions.
(S) 5×10–3 M KOH is a very dilute aqueous solution. For dilute solutions, density ≈ 1 g/ml, so molarity (M) ≈ molality (m). Thus, it matches (2) Molarity = Molality.
Final Answer: P→4, Q→1, R→3, S→2