JEE Main 2025 — Ionic Equilibrium Question with Solution

To determine the percentage dissociation of the salt ${\text{MX}}_3$, consider the following dissociation reaction:

${\text{MX}}_3\to {\text{M}}^{+3}+3{\text{X}}^{-}$

The van't Hoff factor, $\text{i}$, is given as 2. The formula relating $\text{i}$ to the degree of dissociation, $\alpha$, is:

$\text{i}=1+(\text{n}-1)\alpha$

Here, $\text{n}$ represents the total number of ions produced per formula unit of the salt. For ${\text{MX}}_3$, dissociation yields:

1 ${\text{M}}^{+3}$ ion

3 ${\text{X}}^{-}$ ions

Thus, $\text{n}=4$. Substituting into the formula, we have:

$2=1+(4-1)\alpha$

Simplifying, we solve for $\alpha$:

$2=1+3\alpha$

$3\alpha =1$

$\alpha =\frac{1}{3}\approx 0.3333$

As a percentage, this degree of dissociation is:

$\alpha \times 100\%\approx 33.33\%$

Rounding to the nearest integer gives:

$33\%$

Therefore, the percentage dissociation of the salt ${\text{MX}}_3$ is approximately 33%.