For solution of urea (M_w = 60 g/mol) with 6 g in 100 g water, molality:
A1 m
B6 m
C1 m (= 6/60 / 0.1 kg water)
D0.1 m
Answer & Solution
Correct answer: C. 1 m (= 6/60 / 0.1 kg water)
Moles urea = 6/60 = 0.1 mol. Kg water = 0.1. Molality = 0.1/0.1 = 1 mol/kg. ΔT_f = 1.86 × 1 = 1.86 K (freezing pt = -1.86°C).
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