JEE Main 2025 — Ionic Equilibrium Question with Solution

To calculate the percent degree of ionization of the compound ${\mathrm{MX}}_2$, we begin by considering its dissociation process:

${\mathrm{MX}}_2\to {\mathrm{M}}^{+2}+2{\mathrm{X}}^{-}$

Next, we use the van't Hoff factor (i), calculated as the ratio of the normal molar mass to the observed molar mass:

$i=\frac{\text{normal molar mass}}{\text{observed molar mass}}=\frac{164}{65.6}$

Now, expressing the relationship of ionization in terms of the van't Hoff factor, we have:

$1+(3-1)\alpha =\frac{164}{65.6}$

Simplifying this equation gives us:

$2\alpha =\frac{164}{65.6}-1=\frac{98.4}{65.6}$

Solving for $\alpha$:

$\alpha =0.75$

Therefore, the percent dissociation, or the percent degree of ionization, is:

$\text{Percent dissociation}=75\%$