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, \(\Delta T_f\), is determined by the formula: \[\Delta T_f = i \cdot K_f \cdot m\]In this calculation:- \(i\) represents the van’t Hoff factor, which is 2 for KCl due to its dissociation into \( \text{K}^+ \) and \( \text{Cl}^- \).- \(K_f\) is the freezing point depression constant for water, given as 1.86 K kg mol\(^{-1}\).- \(m\) is the molality of the solution, which is 1 mol/kg.Substituting these values, we get:\[\Delta T_f = 2 \cdot 1.86 \cdot 1 = 3.72^\circ C\]Since the normal freezing point of water is \(0^\circ C\), the solution's freezing point will be:\[0^\circ C - 3.72^\circ C = -3.72^\circ C\] Consequently, the correct option is (A).