A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 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

$$DeltaTf=Kfm$$

anonymous · Answer

DeltaTf is  freezing-point depression
Kf is  freezing point constant
m is molality
m=1000W2/M2*W1

anonymous · Answer

That did help a little but I am still confused

anonymous · Answer

W2=10.20 grams of glucose 
M2=180g of glucose
W1=355 grams of water
Kf= -1.86 °C/m

anonymous · Answer

Okay I worked it out and got that the freezing point depression = -0.2969014..... Is that correct? @asmagul

anonymous · Answer

yes u r right

anonymous · Answer

Thank you so much!

anonymous · Answer

@asmagul