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A solution of urea in water has a boiling point of 100.18°C. Calculate the freezing point of the solution. (Kf for water is 1.86 K kg mol-1 and for water is 0.512 K kg mol-1 ). |
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Answer» Kb = 0.512 K kg mol-1 Kf = 1.86 K kg mol-1 ΔTb = 100.18° – 100° = 0.18 ΔT = Kb x molality molality = 0.18/0.512 Now, ΔTf = Kf x molality = 1.86 x \(\frac{0.18}{0.512}\) = 0.6539 Freezing point of solutions = 0 – 0.6539 = -0.6539 |
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