What is the freezing point in °C of a solution prepared by dissolving 6.80 g of MgCl2 in 110 g of water? The value of Kf for water is 1.86 (°C·kg)/mol, and the van't Hoff factor for MgCl2 is i = 2.7.

Calculate the molality of the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent. Use the formula: \( m = \frac{\text{moles of MgCl}_2}{0.110 \text{ kg of water}} \).

Apply 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, and \( m \) is the molality. Substitute the known values: \( i = 2.7 \), \( K_f = 1.86 \text{ °C·kg/mol} \), and the calculated molality.