What is the expected freezing point of a 0.50 m solution of Na₂SO₄ 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

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The expected freezing point depression can be calculated using the formula: ΔTf = Kf*m where ΔTf is the change in freezing point, Kf is the freezing point depression constant (1.86°C/m for water), and m is the molality of the solution. First, we need to find the molality of the solution: molality = moles of solute / kg of solvent The molar mass of Na₂SO₄ is: 22.99 g/mol (Na) + 32.06 g/mol (S) + 192.22 g/mol (B) + 15.99 g/mol (O) = 263.26 g/mol So, we can find the moles of solute in 1 kg of water: 0.50 mol/L x 263.26 g/mol = 131.63 g/L 131.63 g/L / 1000 g/kg = 0.13163 mol/kg Therefore, the molality is: molality = 0.13163 mol/kg Now we can calculate the expected freezing point depression: ΔTf = Kf*m ΔTf = 1.86°C/m * 0.13163 mol/kg = 0.245°C The freezing point depression is negative, which means that the expected freezing point is lower than the freezing point of pure water. We can calculate the expected freezing point as: freezing point = 0°C - ΔTf = 0°C - 0.245°C = -2.45°C Rounding to one decimal place, the expected freezing point is -2.4°C. The closest choice is C, -2.8°C.

What Is the Expected Freezing Point of a 0