For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i) </span>where: ΔTf = change in freezing point = (Ti - Tf) Ti = freezing point of pure water = 0 celsius Tf = freezing point of water with solute = ? Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water) m = molality of solution (mol solute/kg solvent) = ? i = ions in solution = 3
Computing for molality: Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem, ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
