Colligative
properties calculations are used for this type of problem. Calculations are as
follows:
ΔT(freezing point) = -(Kf)m
-20.0 = -1.86 °C kg / mol (m)
m = 10.75 mol/kg
10.75 mol C2H6O2 / kg water ( 1 L water ) ( 1 kg/L ) (62.07 g/mol )= 667.25 g
Hope this answers the question. Have a nice day.
