Predict the freezing point of a solution prepared by dissolving 8.9 g MgCl₂ in 98 g H₂O. Kf 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

First, we need to calculate the molality of the solution using the formula:

molality = moles of solute / mass of solvent in kg

We have 8.9 g of MgCl2 and 98 g of H2O. The molar mass of MgCl2 is 95.211 g/mol.

moles of MgCl2 = 8.9 g / 95.211 g/mol = 0.0935 mol

mass of H2O = 98 g / 1000 = 0.098 kg

molality = 0.0935 mol / 0.098 kg = 0.9534 m

Next, we can use the freezing point depression equation:

∆Tf = Kf x molality

where ∆Tf is the change in freezing point and Kf is the freezing point depression constant for water.

Kf for water is 1.86°C m-1, so:

∆Tf = 1.86°C m-1 x 0.9534 m

∆Tf = 1.774244 °C

The freezing point of pure water is 0°C, so the freezing point of the solution will be:

freezing point = 0°C - ∆Tf

freezing point = 0°C - 1.774244°C

freezing point = -1.774244°C

Therefore, the best answer is D) -5.3 °C (the negative sign indicates a lower freezing point than pure water).

Predict the Freezing Point of a Solution Prepared by Dissolving