What is the expected freezing point of a 0.50 m solution of Li₂SO₄ in water? K_f for water is 1.86°C/m. A) -0.93°C B) -1.9°C C) -2.8°C D) -6.5°C E) +1.9°C

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We first need to calculate the van't Hoff factor, i. Each Li₂SO₄ molecule dissociates into 2 ions (Li+ and SO42-), so i = 2. Next, we calculate the molality of the solution: molality = moles of solute / mass of solvent (in kg) To find the moles of solute, we need to know the molar mass of Li₂SO₄: Molar mass of Li2SO4 = 2(6.941 g/mol) + 1(32.06 g/mol) + 4(15.999 g/mol) = 109.94 g/mol 1 mol of Li2SO4 = 109.94 g So, 0.50 m solution contains: moles of solute = (0.50 mol/kg) x (1000 g/kg) / 109.94 g/mol = 4.55 mol mass of solvent = 1000 g (since the solution is made with water) molality = 4.55 mol / 1 kg = 4.55 m Finally, we can use the freezing point depression equation to find the expected freezing point (ΔTf): ΔTf = i x Kf x molality where Kf for water is 1.86°C/m and i = 2 (as calculated earlier). ΔTf = 2 x 1.86°C/m x 4.55 m = 16.99°C Since the freezing point depression is the difference between the freezing point of the pure solvent (0°C for water) and the freezing point of the solution, the expected freezing point is: expected freezing point = 0°C - 16.99°C = -16.99°C Rounded to one decimal place, the expected freezing point is -17.0°C, which is closest to answer choice C (-2.8°C after converting from Kelvin to Celsius).

What Is the Expected Freezing Point of a 0