The freezing point of a solution made by dissolving 8.9 g of MgCl₂ in 98 g H₂O is -4.3 °C.K_f for water is 1.86 °C m^-1.The van't Hoff i factor for this solution is ________. A)1.2 B)2.7 C)2.4 D)2.0 E)3.0

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

The molar mass of MgCl2 is 95.21 g/mol. Therefore, the number of moles of MgCl2 in 8.9 g is:

moles of MgCl2 = 8.9 g / 95.21 g/mol = 0.0935 mol

Also, the mass of the solvent (water) is:

mass of solvent = 98 g = 0.098 kg

So, the molality of the solution is:

molality = 0.0935 mol / 0.098 kg = 0.952 mol/kg

Next, we need to calculate the expected freezing point depression using the formula:

ΔTf = Kf x molality x i

where Kf is the freezing point depression constant for the solvent (water), and i is the van't Hoff factor. The Kf for water is given as 1.86 °C/m.

Substituting the given values, we get:

-4.3 °C = 1.86 °C/m x 0.952 mol/kg x i

Solving for i, we get:

i = -4.3 °C / (1.86 °C/m x 0.952 mol/kg) = 2.4

Therefore, the van't Hoff i factor for this solution is 2.4, and the correct answer is (C).

The Freezing Point of a Solution Made by Dissolving 8