Question

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.

Answer 1

Dudes, please help. I really need some assistance.

Answer 2

Use Number of moles=Mass(g)/Molar Mass(g/mol)
10.20/(12*6)+(1*12)+(16*6)=10.20/180=0.056667 moles
Number of moles=Concentration(mol/dm^3)*Volume(dm^3)
1 gram is equivalent to 1 cm^3, change 355 grams to cm^3. Convert cm^3 unit to dm^3
by dividing by 1000.
Glucose does not ionize(break into ions) its a non electrolyte, it can dissolve well in water only.
non-electrolytes dissolved in water, the van' t Hoff factor is essentially 1.