Five (5.00) grams of glucose, C6H12)6, is dissolved in 500.0 grams of acetic acid. What is the new freezing point and boiling point for the solution?

Question

Kf acetic acid = 3.90, Kb acetic acid = 3.07
(normal freezing point for acetic acid = 16.60 dg C, boiling point = 118.5 dg C)

f.p. = ??????
b.p. = ??????

b.p: delta T = Kb*molality = 3.07*m
f.p: delta T = Kf*molality = 3.90*m
molality = mols/kg solvent
mols = g/molar mass
Post your work if you get stuck.