A solution of urea boils at 100. 18 degree Celsius at atmospheric pressure. If kf of water is1.86 andkb is 0.512. What is the freezing pointof soln

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$B o i l i n g p o i n t o f s o l u t i o n = 100.18 ° C K_{b} = 0.512 K k g m o l^{- 1} K_{f} = 1.86 K k g m o l^{- 1} ∆ T_{b} = 100.18 - 100 = 0 . 18^{°} C ∆ T_{b} = K_{b} \times m o l a l i t y T h u s, m o l a l i t y = \frac{0.18}{0.512} ∆ T_{f} = K_{f} \times m o l a l i t y = 1.86 \times \frac{0.18}{0.512} = 0.6539 T h u s, t h e f r e e z i n g p o i n t o f t h e s o l u t i o n = 0 - 0.6539 = - 0 . 6539^{°} C$

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A solution of urea boils at 100. 18 degree Celsius at atmospheric pressure. If kf of water is1.86 andkb is 0.512. What is the freezing pointof soln

Dear student, Boiling point of solution = 100.18 °CKb = 0.512 K kg mol-1Kf = 1.86 K kg mol-1∆Tb = 100.18 - 100 = 0.18°C∆Tb = Kb × molalityThus, molality = 0.180.512∆Tf = Kf × molality = 1.86 ×0.180.512= 0.6539Thus, the freezing point of the solution = 0 - 0.6539 = -0.6539°C Regards