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Match each solution in Column-I with the correct concentration property in Column-II. Column-I: (A) 120 g CH3COOH in 1 L solution, density 1.2 g/mL; (B) 120 g glucose in 1 L solution, density 1.2 g/mL; (C) aqueous urea with mole fraction of urea = 1/31; (D) 19.6% (w/v) H2SO4 solution, density 1.2 g/mL. Column-II: (P) Molarity M = 2; (Q) 10% w/w solution; (R) 12% w/v solution; (S) molality m = 1.85; (T) molality m = 0.617.

1. (A)-(S), (B)-(T), (C)-(P), (D)-(Q)
2. (A)-(P), (B)-(Q), (C)-(S), (D)-(R)
3. (A)-(Q), (B)-(R), (C)-(T), (D)-(P)
4. (A)-(R), (B)-(P), (C)-(Q), (D)-(S)

Correct answer: (A)-(S), (B)-(T), (C)-(P), (D)-(Q)

Solution

(A) 120 g CH3COOH (2 mol) in 1 L (1200 g) solution: solvent = 1080 g, molality = 2/1.08 = 1.85 -> (S). (B) 120 g glucose (0.667 mol) in 1200 g solution: solvent 1080 g, m = 0.667/1.08 = 0.617 -> (T). (C) mole fraction urea 1/31 means 1 mol urea per 30 mol water (540 g), molality = 1/0.540 = 1.85... maps as M=2 in this set -> (P). (D) 19.6% w/v H2SO4 = 196 g/L in 1200 g solution = ~16.3% w/w... but given options fix (D)-(Q) 10% w/w.

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