The percentage dissociation of a salt $\left(\mathrm{MX}_3\right)$ solution at given temperature (van't Hoff factor $\mathrm{i}=2$ ) is ___________ %(Nearest integer)

<p>To determine the percentage dissociation of the salt $\text{MX}_3$, consider the following dissociation reaction:</p> <p>$ \text{MX}_3 \rightarrow \text{M}^{+3} + 3\text{X}^{-} $</p> <p>The van't Hoff factor, $\text{i}$, is given as 2. The formula relating $\text{i}$ to the degree of dissociation, $\alpha$, is:</p> <p>$ \text{i} = 1 + (\text{n} - 1) \alpha $</p> <p>Here, $\text{n}$ represents the total number of ions produced per formula unit of the salt. For $\text{MX}_3$, dissociation yields:</p> <p><p>1 $\text{M}^{+3}$ ion</p></p> <p><p>3 $\text{X}^{-}$ ions</p></p> <p>Thus, $\text{n} = 4$. Substituting into the formula, we have:</p> <p>$ 2 = 1 + (4 - 1) \alpha $</p> <p>Simplifying, we solve for $\alpha$:</p> <p>$ 2 = 1 + 3\alpha $</p> <p>$ 3\alpha = 1 $</p> <p>$ \alpha = \frac{1}{3} \approx 0.3333 $</p> <p>As a percentage, this degree of dissociation is:</p> <p>$ \alpha \times 100\% \approx 33.33\% $</p> <p>Rounding to the nearest integer gives:</p> <p>$ 33\% $</p> <p>Therefore, the percentage dissociation of the salt $\text{MX}_3$ is approximately 33%.</p>