Final answer:
The expected freezing point of a 0.50 m solution of Li₂SO₄ in water is calculated to be -2.79°C using the formula for freezing point depression and accounting for the dissociation of Li₂SO₄ into three ions in solution.
Step-by-step explanation:
The question asks about the freezing point depression of a 0.50 m solution of Li₂SO₄ in water. We know that Li₂SO₄ will dissociate into three ions - two Li+ and one SO₄²⁻. Therefore, we will have a total of 3 moles of ions for every 1 mole of Li₂SO₄. Using the formula ΔTf = i × Kf × m (where ΔTf is the freezing point depression, i is the van't Hoff factor, Kf is the freezing point depression constant, and m is the molality), and knowing that the Kf for water is 1.86 °C/m, we can calculate the freezing point depression as follows:
ΔTf = 3 × 1.86 °C/m × 0.50 m = 2.79 °C.
Since the normal freezing point of water is 0.0°C, subtracting the depression from 0.0°C gives the new freezing point: 0.0°C - 2.79°C = -2.79°C. Thus, the expected freezing point of a 0.50 m solution of Li2SO4 in water is -2.79°C.