question_answer 32)

A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Answer:

Tf  = 273.15 ? 271.00 = 2.15 K Now Tf  = Kf x m           ... (i) But   Substituting the values in equation (i), we get 2.15 = Kf x  ?.. (ii) Also, Tf = Kf x       ?.. (iii)   Dividing (iii) by (ii), we get  = = 273.15 ?T?f Freezing point of solution = 269.07k.