Asked by Anonymous
What is the freezing point in degrees Celsius of a solution prepared by dissolving 11.3 (g) of calcium nitrate in 115 (g) of water? K of f for water is 1.86 degrees Celsius/m
Answers
Answered by
DrBob222
mols Ca(NO3)2 = grams/molar mass.
molality = mols Ca(NO3)2/kg solvent
delta T = i*Kf*m
i = 3 for Ca(NO3)2
Then 0C-delta T = freezing point.
molality = mols Ca(NO3)2/kg solvent
delta T = i*Kf*m
i = 3 for Ca(NO3)2
Then 0C-delta T = freezing point.
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