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

ΔT

_{f }=K_{f}bibut 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 MgCl

_{2}to do this.Mg is 24.305 amu

Cl is 35.34 amu

so MgCl

_{2}24.305+2(35.34)= 94.985 g/mol and we have 4.00 g =0.0421119124072221929778386060957 mol

94.985 g/mol

and the molality is

4.2111x10

^{-2}mol=0.3828355673383835725258055099609 m 0.110 kg

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.