What is the freezing point of a 2.12 molal solution of NaCl in water if the Kf is -1.86 ?

To determine the freezing point of a 2.12 molal (m) solution of NaCl in water, we use the freezing point depression formula: \[ \Delta T_f = i \times K_f \times m \] Where: - \(\Delta T_f\) = change in freezing point - \(i\) = van 't Hoff factor (number of particles the solute dissociates into) - \(K_f\) = cryoscopic constant (freezing point depression constant) - \(m\) = molality of the solution **Given:** - \(m = 2.12\) mol/kg - \(K_f = 1.86\) °C·kg/mol (Note: Freezing point depression constants are typically positive, even if the question provides a negative value, because the formula accounts for the depression itself.) - For NaCl, \(i = 2\) (since NaCl dissociates into Na\(^+\) and Cl\(^-\)) **Calculation:** \[ \Delta T_f = 2 \times 1.86 \times 2.12 = 7.89 \text{ °C} \] This means the freezing point is lowered by 7.89°C from that of pure water. **Final Freezing Point:** \[ \text{Freezing point} = 0°C - 7.89°C = -7.89°C \] **Rounded to one decimal place:** \[ \text{Freezing point} \approx -7.9°C \] **Answer:** The solution freezes at approximately –7.9 °C.