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

813 views

closed

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 273.15K.

+1 vote
by (47.9k points)
selected by

Molar mass of cane sugar

C12H22O11 = 342 g mol-1

Molality of sugar = $\frac{5\times1000}{342\times 100}$= 0.146

∆T2 for sugar solution = 273.15 – 271 = 2.15°

∆Tf = Kf x m

Kf = $\frac{2.15}{0.146}$

Molality of glucose solution = $\frac{5}{180}$ x $\frac{1000}{100}$= 0.278 m

∆Tf (Glucose) = $\frac{2.15}{0.146}$ x 0.278 = 4.09°K

Freezing point of glucose solution = 273.15 – 4.09 = 269.06 K