The observed and normal molar masses of compound $\mathrm{MX}_2$ are 65.6 and 164 respectively. The percent degree of ionisation of $\mathrm{MX}_2$ is __________%. (Nearest integer)

<p>To calculate the percent degree of ionization of the compound $\mathrm{MX}_2$, we begin by considering its dissociation process:</p> <p>$ \mathrm{MX}_2 \rightarrow \mathrm{M}^{+2} + 2\mathrm{X}^{-} $</p> <p>Next, we use the van't Hoff factor (i), calculated as the ratio of the normal molar mass to the observed molar mass:</p> <p>$ i = \frac{\text{normal molar mass}}{\text{observed molar mass}} = \frac{164}{65.6} $</p> <p>Now, expressing the relationship of ionization in terms of the van't Hoff factor, we have:</p> <p>$ 1 + (3 - 1)\alpha = \frac{164}{65.6} $</p> <p>Simplifying this equation gives us:</p> <p>$ 2\alpha = \frac{164}{65.6} - 1 = \frac{98.4}{65.6} $</p> <p>Solving for $\alpha$:</p> <p>$ \alpha = 0.75 $</p> <p>Therefore, the percent dissociation, or the percent degree of ionization, is:</p> <p>$ \text{Percent dissociation} = 75\% $</p>