Question

Calculate the percent dissociation of \(0.02\text{ m}\) solution if its freezing point depression is \(0.046\text{ K}\) . \([\text{K}_\text{f} \text{ for water } = 1.86\text{ K kg mol}^{-1}; \text{n} = 2]\)

• A. \(12.3%\)
• B. \(23.6%\)
• C. \(35.00%\)
• D. \(48.1%\)

Hint:

For dissociation problems: First find \(i\), then use \(i = 1 + \alpha(n-1)\). For \(n=2\), simply: \(i = 1 + \alpha\).

Answer 1

The Correct Option is B

Concept:
Freezing point depression is given by: \[ \Delta T_f = i K_f m \] where \(i\) is the van't Hoff factor. Step 1: Calculate van't Hoff factor \(i\). \[ i = \frac{\Delta T_f}{K_f m} = \frac{0.046}{1.86 \times 0.02} \] \[ i = \frac{0.046}{0.0372} \approx 1.24 \]
Step 2: Use relation between \(i\) and degree of dissociation. For \(n = 2\): \[ i = 1 + \alpha(n-1) \] \[ 1.24 = 1 + \alpha(1) \] \[ \alpha = 0.24 \]
Step 3: Convert into percentage. \[ %\text{ dissociation} = 0.24 \times 100 = 24% \approx 23.6% \]
Step 4: Conclusion. \[ {23.6%} \]