what is the expected freezing point of a solution containing of 18.8 grams of mgcl2 dissolved 200.00. grams of water?

Molar mass of MgCl₂ = 24.3+ 35.5×2 = 95.3 g/mol

Each mole of MgCl₂ contains 1 mole of Mg²⁺ ions and 2 moles of Cl⁻ ions, i.e. 3 moles of ions.
Hence, van't Hoff factor, i = 3
Molality of the solution, m = (18.8/95.3 mol) / (200.00/1000 kg water) = 0.9864 m
Freezing point depression constant, Kf = 1.86 °C/m

Freezing point depression, ΔTf = i Kf m = 3 × 1.86 × 0.9864 °C = 5.50°C

Normal freezing point of water = 0.00°C
Freezing point of the solution = (0.00 - 5.50) °C = -5.50 °C