Question

What is the molality of an aqueous solution of KBr having freezing point -3.72\(^\circ\)C (K\(_f\) for water is 1.86 K kg mol\(^{-1}\))?

Answer 1

Step 1: Recall the formula for freezing point depression.

\[ \Delta T_f = i \cdot K_f \cdot m \] where:

• \(\Delta T_f\) = depression in freezing point
• \(i\) = van’t Hoff factor (number of particles formed per formula unit)
• \(K_f\) = cryoscopic constant (for water = 1.86 K·kg·mol\(^{-1}\))
• \(m\) = molality

Step 2: Calculate \(\Delta T_f\).

The freezing point of pure water = \(0^\circ C\). Given freezing point of solution = \(-3.72^\circ C\). \[ \Delta T_f = 0 - (-3.72) = 3.72 \, K \]

Step 3: Value of \(i\) for KBr.

KBr dissociates completely in water as: \[ \text{KBr} \;\;\rightarrow\;\; \text{K}^+ + \text{Br}^- \] So, \(i = 2\).

Step 4: Substitute values in the formula.

\[ \Delta T_f = i \cdot K_f \cdot m \] \[ 3.72 = 2 \times 1.86 \times m \]

Step 5: Solve for molality.

\[ m = \frac{3.72}{2 \times 1.86} = \frac{3.72}{3.72} = 1.0 \, \text{mol kg}^{-1} \]

Final Answer:

The molality of the solution is \[ \boxed{1.0 \, \text{mol kg}^{-1}} \]