The molal freezing point constant for water is -1.86°C/m.At what temperature will a solution containing 8.27 g CaCl₂ and 45.0 g H₂O begin to freeze? Assume that no ion-pairing occurs between Ca²⁺ and Cl^-.

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The freezing point depression of a solution can be calculated using the formula ΔT_f = i * K_f * m, where ΔT_f is the change in freezing point, i is the van't Hoff factor (which represents the number of particles the solute dissociates into), K_f is the molal freezing point depression constant for the solvent (in this case, water, which is -1.86°C/m), and m is the molality of the solution.For CaCl2:- The molar mass of CaCl2 is approximately 40.08 (Ca) + 35.45*2 (Cl) = 110.98 g/mol.- The amount of CaCl2 in moles is 8.27 g / 110.98 g/mol = 0.0745 mol.- The van't Hoff factor (i) for CaCl2, which dissociates into 3 ions (Ca^2+ and 2 Cl^-), is 3.- The mass of water in kilograms is 45.0 g / 1000 = 0.045 kg.- The molality (m) of the solution is 0.0745 mol / 0.045 kg = 1.655 mol/kg.Using the formula:ΔT_f = i * K_f * m = 3 * (-1.86°C/m) * 1.655 mol/kg = -9.24°C.This means the solution will begin to freeze at 0°C - 9.24°C = -9.24°C.

The Molal Freezing Point Constant for Water Is -1