Question

The normal freezing point of water is 0.00°C. Calculate the expected freezing point of a water solution prepared by dissolving 57.3 g of MgCl2 (molar mass: 95.211 g/mol) in 350.4 g of water. Assume that the van't Hoff factor is equal to the number of ions. Kf = 1.86°C/m for water.

Answer 1

Answer: The Answer is -1.395°C.

Explanation: To calculate the expected freezing point of a water solution, we can use the following equation:

ΔTf = i * Kf * m

where:

ΔTf is the change in freezing point (°C)

i is the van't Hoff factor

Kf is the freezing point depression constant of water (1.86°C/m)

m is the molality of the solution (mol/kg)

First, we need to calculate the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. To calculate the molality, we need to know the mass of solute (MgCl2) and the mass of solvent (water).

molality of solution = moles of solute / kg of solvent

molality of solution = 57.3 g / (95.211 g/mol) / (350.4 g / 1000 g/kg)

molality of solution = 0.25 m

Next, we need to calculate the van't Hoff factor. The van't Hoff factor is a measure of how many ions a solute dissociates into in solution. MgCl2 dissociates into two ions: Mg2+ and 2Cl-. Therefore, the van't Hoff factor for MgCl2 is 3.

Now that we have the molality and the van't Hoff factor, we can calculate the change in freezing point.

ΔTf = i * Kf * m

ΔTf = 3 * 1.86°C/m * 0.25 m

ΔTf = 1.395°C

The expected freezing point of the solution is therefore 0.00°C - 1.395°C = -1.395°C.

Therefore, the answer is -1.395°C.