Question

A compound having molecular formula \( \text{MX}_3 \) has van't Hoff factor of 2. What is the degree of dissociation?

Hint:

The van't Hoff factor \( i \) is related to the degree of dissociation \( \alpha \) by the formula \( i = 1 + \alpha (n - 1) \), where \( n \) is the number of ions produced.

Answer 1

The van't Hoff factor \( i \) is related to the degree of dissociation \( \alpha \) by the equation: \[ i = 1 + \alpha (n - 1) \] where \( n \) is the number of ions into which the compound dissociates. For \( \text{MX}_3 \), \( n = 4 \) (since it dissociates into \( \text{M}^{4+} \) and 3 \( \text{X}^- \)). Given that \( i = 2 \), we can substitute into the equation: \[ 2 = 1 + \alpha (4 - 1) \] \[ 2 = 1 + 3\alpha \] \[ \alpha = \frac{1}{3} \] Thus, the degree of dissociation is \( \alpha = \frac{1}{3} \). Therefore, the correct answer is 0.3.