An aqueous solution of a salt $$MX_2$$ at certain temperature has a Van't Hoff factor of 2. What is the degree of dissociation for this solution of the salt?

First, let us write the dissociation equation for the salt $$MX_2$$ when it is dissolved in water.

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

From this equation we see that one formula unit of $$MX_2$$ produces one cation $$M^{2+}$$ and two anions $$X^-$$, giving a total of $$1 + 2 = 3$$ ions. Thus, the number of particles after complete dissociation, which we denote by $$n$$, is $$n = 3$$.

Now we recall the relation that connects the Van’t Hoff factor $$i$$, the degree of dissociation $$\alpha$$, and the total number of ions produced on full dissociation $$n$$.

$$i = 1 + (n - 1)\alpha$$

This formula states that, starting from one initial particle, complete dissociation of a fraction $$\alpha$$ of those particles increases the total particle count by $$(n - 1)$$ for each dissociating unit.

In the present question, the Van’t Hoff factor is given as $$i = 2$$ and we have already found $$n = 3$$. Substituting these values into the formula, we get

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

Simplify the expression inside the parentheses:

$$2 = 1 + 2\alpha$$

Now isolate $$\alpha$$ by subtracting 1 from both sides:

$$2 - 1 = 2\alpha$$

So this becomes

$$1 = 2\alpha$$

Finally, divide both sides by 2 to solve for $$\alpha$$:

$$\alpha = \dfrac{1}{2} = 0.50$$

Hence, the degree of dissociation of the salt $$MX_2$$ under the stated conditions is $$0.50$$.

Hence, the correct answer is Option A.