Predict the freezing point of a solution prepared by dissolving 8.9 g MgCl₂ in 98 g H₂O.K_f for water is 1.86 °C m^-1. A)3.5 °C B)-3.5 °C C)5.3 °C D)-5.3 °C E)-0.51 °C

Question

Quizplus · Accepted Answer

Firstly, we need to calculate the molality of the solution:
molality = moles of solute / mass of solvent in kg

The molar mass of MgCl2 is 95.21 g/mol, so the number of moles of MgCl2 is:
8.9 g / 95.21 g/mol = 0.0935 mol

The mass of the solvent (water) is 98 g or 0.098 kg.

molality = 0.0935 mol / 0.098 kg = 0.954 mol/kg

Using the freezing point depression equation:
ΔTf = Kf x molality

Given that Kf for water is 1.86 °C/m, we can calculate the freezing point depression:
ΔTf = 1.86 °C/m x 0.954 mol/kg = 1.7756 °C

The freezing point of the solution will be lower than that of the pure solvent (water) by this amount:
Freezing point = 0 °C - 1.7756 °C = -1.7756 °C

Therefore, the predicted freezing point of the solution is -1.7756 °C, which is closest to answer choice D) -5.3 °C.

Predict the Freezing Point of a Solution Prepared by Dissolving