the freezing pointof a 0.05 molal solution of a non electrolyte in water is
Asked by sarvaterahul | 1st Jul, 2016, 08:25: AM
Expert Answer:
∆Tf = kf x m = 1.86 x 0.05 = 0.093°C
Freezing point = 0-0.093 = -0.093°C
∆Tf = kf x m = 1.86 x 0.05 = 0.093°C
Freezing point = 0-0.093 = -0.093°C
Answered by Vaibhav Chavan | 1st Jul, 2016, 09:18: AM
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