Question

The \(\Delta T_b\) value for 0.01 m KCl solution is 0.01 K. What is the Van’t Hoff factor? (Kb for water = 0.52 K kg mol\(^{-1}\))

• A. 1.92
• B. 1.72
• C. 0.96
• D. 0.86

Hint:

The Van’t Hoff factor (i) represents the number of particles formed in solution. For KCl, it dissociates into 2 ions, so \(i = 2\) ideally. However, in this case, the calculation shows the effective dissociation.

Answer 1

The Correct Option is A

We know that: \[ \Delta T_b = i \cdot K_b \cdot m \] Where:
- \(\Delta T_b = 0.01 \, K\),
- \(K_b = 0.52 \, \text{K kg mol}^{-1}\),
- \(m = 0.01 \, \text{mol/kg}\). Substitute the values: \[ 0.01 = i \cdot 0.52 \cdot 0.01 \quad \Rightarrow \quad i = \frac{0.01}{0.52 \cdot 0.01} = 1.92 \] Thus, the Van’t Hoff factor is 1.92.