1. Use the relation between normal molar mass and observed molar mass

For an electrolyte, if dissociation occurs, the observed molar mass decreases because the number of particles increases.

The van’t Hoff factor is

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

Given:

$M_{\text{normal}} = 164, \qquad M_{\text{observed}} = 65.6$

So,

$i = \frac{164}{65.6} = 2.5$

2. Write dissociation of $\mathrm{MX_2}$

$\mathrm{MX_2 \rightarrow M^{2+} + 2X^-}$

If degree of ionisation is $\alpha$, then:

• 1 mole of $\mathrm{MX_2}$ gives 3 ions on complete dissociation.
• Total number of particles after dissociation:

$1-\alpha + 3\alpha = 1 + 2\alpha$

Hence,

$i = 1 + 2\alpha$

3. Substitute $i = 2.5$

$2.5 = 1 + 2\alpha$

$2\alpha = 1.5$

$\alpha = 0.75$

4. Convert to percent ionisation

$\%\text{ ionisation} = 0.75 \times 100 = 75\%$

5. Final Answer

The percent degree of ionisation is

$\boxed{75}$