3.2 grams of a compound with a mole weight of 96 g/mol dissolved in 50 grams of water gave a solution that freezes at -1.50°C. Kf for water is 1.86°C/molal. The van't Hoff i factor for this solution is ________.

Question

Quizplus · Accepted Answer

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

molality = moles of solute / mass of solvent in kg

mass of solvent = 50 g / 1000 = 0.05 kg

moles of solute = 3.2 g / 96 g/mol = 0.03333 mol

molality = 0.03333 mol / 0.05 kg = 0.6666 mol/kg

Next, we can use the freezing point depression equation:

ΔTf = Kf * molality * i

where ΔTf is the change in freezing point, Kf is the freezing point depression constant, molality is the molality of the solution, and i is the van't Hoff factor.

Rearranging the equation to solve for i:

i = ΔTf / (Kf * molality)

i = (-1.50°C) / (1.86°C/molal * 0.6666 mol/kg)

i ≈ 1.2

Therefore, the van't Hoff i factor for this solution is approximately 1.2.

The answer is B.

32 Grams of a Compound with a Mole Weight of 96