What is the freezing point in °C of a solution prepared by dissolving 4.00g of MgCl2 in 110 g of water? The value of Kf for water is 1.86 (°C⋅kg)/mol, and the van't Hoff factor for MgCl2 is i = 2.7
freezing point depression is defined as
but to use that equation we need the molality of the solution which is moles of solute in kilograms of solvent. We need to calculate the molecular weight of MgCl2 to do this.
Mg is 24.305 amu
Cl is 35.34 amu
so MgCl2 24.305+2(35.34)= 94.985 g/mol and we have
4.00 g =0.0421119124072221929778386060957 mol
and the molality is
4.2111x10-2 mol=0.3828355673383835725258055099609 m
and the finally the freezing point depression:
ΔTf =(1.86)(0.38283)(2.7)= 1.9ºC
and the actual freezing point would be
0ºC - 1.9ºC = -1.9ºC
be careful when you read these problems to determine if they are asking for the actual freezing temperature of the solution or how much the solute changes it. In this case they are nearly the same but that's because we were working with water.