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

$ \text{MX}_3 \rightarrow \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%.