A molecule M associates in a given solvent according to the equation $$M \rightleftharpoons (M)_n$$. For a certain concentration of M, the van't Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is:

The association reaction is given by the equation $$ M \rightleftharpoons (M)_n $$, where $$ n $$ molecules of $$ M $$ associate to form one molecule of $$ (M)_n $$. The van't Hoff factor $$ i $$ is 0.9, and the fraction of associated molecules is 0.2. We need to find the value of $$ n $$.

The van't Hoff factor $$ i $$ is defined as the ratio of the total number of particles after association to the total number of particles before association. Let the initial number of moles of $$ M $$ be 1 for simplicity. Before association, the total number of moles of particles is 1.

Let $$ \alpha $$ be the fraction of the initial molecules that have associated. Here, $$ \alpha = 0.2 $$. When $$ \alpha $$ moles of $$ M $$ associate, they form $$ \frac{\alpha}{n} $$ moles of $$ (M)_n $$, since $$ n $$ molecules combine to form one complex. The remaining unassociated moles of $$ M $$ are $$ 1 - \alpha $$. Therefore, the total moles of particles after association is $$ (1 - \alpha) + \frac{\alpha}{n} $$.

The van't Hoff factor is given by:

$$ i = \frac{(1 - \alpha) + \frac{\alpha}{n}}{1} = 1 - \alpha + \frac{\alpha}{n} $$

Substituting the given values $$ i = 0.9 $$ and $$ \alpha = 0.2 $$:

$$ 0.9 = 1 - 0.2 + \frac{0.2}{n} $$

$$ 0.9 = 0.8 + \frac{0.2}{n} $$

Subtract 0.8 from both sides:

$$ 0.9 - 0.8 = \frac{0.2}{n} $$

$$ 0.1 = \frac{0.2}{n} $$

Solve for $$ n $$:

$$ n = \frac{0.2}{0.1} = 2 $$

This calculation shows that $$ n = 2 $$ satisfies the given conditions. The fraction of associated molecules is interpreted as the fraction of the initial molecules that have associated, which is $$ \alpha = 0.2 $$, and the van't Hoff factor $$ i = 0.9 $$ is consistent with $$ n = 2 $$.

Hence, the correct answer is Option C.