Question

what is the freezing point of a solution that contains 250g of sucrose, c12h22o11, in 500g of water? water has a freezing point depression of -1.86 degrees celcious/m.

Answer 1

you must calculate the molal sucrose
molal = (mass/Mr) x (1000/mass of water)
molal = (250/342)x (1000/500)= 1.46
\[\Delta Tf = molal \times Kf\] = 1.46 x (-1.86) = -2.72
Freezing point depression = -2.72 degree Celcius