Two salts $$A_2X$$ and MX have the same value of solubility product of $$4.0 \times 10^{-12}$$. The ratio of their molar solubilities i.e. $$\frac{S(A_2X)}{S(MX)}$$ = ________. (Round off to the Nearest Integer).

We are given that salts $$A_2X$$ and $$MX$$ have the same solubility product $$K_{sp} = 4.0 \times 10^{-12}$$. We need to find the ratio of their molar solubilities.

For salt $$A_2X$$, it dissociates as $$A_2X \rightleftharpoons 2A^+ + X^{2-}$$. If the molar solubility is $$S_1$$, then $$[A^+] = 2S_1$$ and $$[X^{2-}] = S_1$$. Therefore, $$K_{sp} = (2S_1)^2 \times S_1 = 4S_1^3$$.

So $$4S_1^3 = 4.0 \times 10^{-12}$$, which gives $$S_1^3 = 1.0 \times 10^{-12}$$, hence $$S_1 = 10^{-4}$$ mol/L.

For salt $$MX$$, it dissociates as $$MX \rightleftharpoons M^+ + X^-$$. If the molar solubility is $$S_2$$, then $$K_{sp} = S_2 \times S_2 = S_2^2$$.

So $$S_2^2 = 4.0 \times 10^{-12}$$, which gives $$S_2 = 2.0 \times 10^{-6}$$ mol/L.

The ratio of molar solubilities is $$\frac{S_1}{S_2} = \frac{10^{-4}}{2.0 \times 10^{-6}} = \frac{10^{-4}}{2 \times 10^{-6}} = 50$$.

The answer is $$\mathbf{50}$$.