Question

What is the expected freezing point of a 0.50 m solution of Li2SO4 in water? Kf for water is 1.86∘C kg mol 1

A) -0.93 ∘C
B) 1.9 ∘C
C) -1.9 ∘C
D) -6.5 ∘C
E) -2.8 ∘C

Answer 1

The expected freezing point of a 0.50 m solution of Li2SO4 in water is calculated using the freezing point depression formula, and is found to be -2.8°C.

Step-by-step explanation:

The question involves finding the expected freezing point of a 0.50 m solution of Li2SO4 in water by using the concept of freezing point depression in solutions. This concept states that the freezing point of a solvent decreases when a solute is dissolved in it. For the given solution of lithium sulfate (Li2SO4), we need to consider the number of ions it produces, since Li2SO4 will dissociate into three ions (2 Li+ and 1 SO42-). Therefore, the molality gets effectively multiplied by the number of particles formed upon dissociation. Using the freezing point depression formula ΔTf = Kf x m x i, where ΔTf is the freezing point depression, Kf is the molal freezing-point depression constant for water (which is 1.86°C·kg/mol), m is the molality of the solution, and i is the van't Hoff factor (which is 3 for Li2SO4).

Now, we calculate ΔTf = (1.86 °C·kg/mol) x (0.50 m) x (3) = 2.79 °C. Since freezing point depression is a decrease in temperature, the freezing point will be -2.79°C, which we round to -2.8°C (answer E).