A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 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.

Question

anonymous · Answer

Calculate moles first:
moles glucose = 15.5 g/ atomic mass = 15.5 g/ (6*12+12*1+6*16) = 15.5/180 = 0.086
Calculate molality next: (molality and molarity are the same when water is the solvent)
0.086 moles/ .245 kg water = 0.3515 m
Calculate temp depression:
-1.86 oC/M *.3515 m = 0.65 oC
so the freezing point will drop from 0oC to -0.65oC.

anonymous · Answer

i was so lost, but i get it more know