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A solution of urea (mol. mass 56 g mol⁻¹) boils at 100.18°C at the atmospheric pressure. If Kb and Kf for water are 1.86 and 0.512 K kg mol⁻¹ respectively, the above solution will freeze at

1. 0.654°C
2. -0.654°C
3. 6.54°C
4. -6.54°C

Correct answer: -0.654°C

Solution

The boiling point elevation (ΔTb) is 0.18°C, and using ΔTb = Kb × m, we calculate the molality (m) as 0.18 / 0.512 = 0.351 mol kg⁻¹. For freezing point depression, ΔTf = Kf × m = 1.86 × 0.351 = 0.654°C. Since freezing point decreases, the solution freezes at -0.654°C.

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