What is the freezing point (in degrees Celsius) of 3.28 kg of water if it contains 170.5 g of CaBr2? The freezing point depression constant for water is 1.86 °C/m and the molar mass of CaBr2 is 199.89 g/mol.

Calculate the van't Hoff factor (i) for CaBr2. Since CaBr2 dissociates into three ions (Ca\(^{2+}\) and 2 Br\(^{-}\)), the van't Hoff factor is 3.

Use the freezing point depression formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant (1.86 °C/m for water), and \( m \) is the molality. Substitute the values to find \( \Delta T_f \).