What is the freezing point of a solution of 8.00 g MgCl₂ in 100 g of water? K_f for water is 1.86°C/m for water. A)1.56°C B)-1.56°C C)4.69°C D)-4.69°C

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution:
molality = moles of solute / mass of solvent in kg
We have 8.00 g of MgCl2 and 100 g of water, which is 0.100 kg:
moles of MgCl2 = 8.00 g / (95.211 g/mol) = 0.084 mol
molality = 0.084 mol / 0.100 kg = 0.84 mol/kg

Next, we can use the freezing point depression equation to calculate the freezing point of the solution:
ΔTf = Kf * molality
ΔTf = 1.86°C/m * 0.84 mol/kg = 1.56°C

Since the freezing point of a solution is lower than the freezing point of the pure solvent, the answer should be negative. Therefore, the correct answer is D.

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