Question

The freezing point of one molal KCl solution, assuming KCl to be completely dissociated in water, is: (\(K_f\) for water = 1.86 K kg mol\(^{-1}\))

• A. \(-3.72^\circ C\)
• B. \(+3.72^\circ C\)
• C. \(-1.86^\circ C\)
• D. \(+2.72^\circ C\)

Hint:

To calculate freezing point depression, remember to use the van’t Hoff factor for dissociation and apply the formula \(\Delta T_f = i \cdot K_f \cdot m\).

Answer 1

The Correct Option is A

The freezing point depression can be calculated using the formula: \[ \Delta T_f = i \cdot K_f \cdot m \] where: - \(i\) is the van’t Hoff factor (which is 2 for KCl, as it dissociates into \( \text{K}^+ \) and \( \text{Cl}^- \)), - \(K_f\) is the freezing point depression constant for water (1.86 K kg mol\(^{-1}\)), - \(m\) is the molality of the solution (1 mol/kg). Thus, \[ \Delta T_f = 2 \cdot 1.86 \cdot 1 = 3.72^\circ C \] The freezing point of water is \(0^\circ C\), so the new freezing point is: \[ 0^\circ C - 3.72^\circ C = -3.72^\circ C \] Therefore, the correct answer is (A).