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.