posted by Henry on .
What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 in 115 g of water? The molal freezing point depression constant for water is 1.86°C/m. (Note that when Ca(NO3)2 dissolves in water Ca2+ and NO32- ions are produced).
mols Ca(NO3)2 = grams/molar mass
molality = mols/kg H2O
delta T = i*Kf*m
You have Kf and m. i for Ca(NO3)2 is 3.
For freezing point, 0-delta T = f.p.
should i get -3.34? can't help but feel as though i did wrong
-3.34 C agrees with my quick calculation.