Question

A solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Answer 1

Change in freezing point=Kf * m.
m- molality of glucose.
m= no. of moles of glucose/mass of solvent(water) in kg.
no. of moles of glucose = 21.5/180.15= 0.1193
m= 0.1193/0.255 = 0.468 moles/kg.
Change in freezing point = (-1.86)*(0.468) = -0.8705 °C

Actually the freezing point increases by 0.8705 °C .