The freezing point (in °C) of a solution containing 0.1g of K3[Fe(CN)6] (Mol. Wt. 329) in 100g of water (Kf = 1.86 K kg mol-1) is:

K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3-

Number of moles of K3[Fe(CN)6] = 0.1g / 329 g/mol = 3.04 × 10-4 moles

Since K3[Fe(CN)6] dissociates into 4 ions, the total number of moles of particles = 4 × 3.04 × 10-4 moles = 1.216 × 10-3 moles

Molality of the solution = (moles of solute) / (mass of solvent in kg) = (1.216 × 10-3 moles) / (0.1 kg) = 0.01216 mol/kg

ΔTf = Kf × m = 1.86 K kg mol-1 × 0.01216 mol/kg = 0.0226 K

Therefore, the freezing point depression is 0.0226 °C.

Since the freezing point of pure water is 0°C, the freezing point of the solution is 0°C - 0.0226°C = -0.0226°C ≈ -2.3 × 10-2 °C