$$M_3A_2$$ is a sparingly soluble salt of molar mass $$y$$ g mol$$^{-1}$$ and solubility $$x$$ g L$$^{-1}$$. The ratio of the molar concentration of the anion ($$A^{3-}$$) to the solubility product of the salt is

Step 1: Find Molar Solubility (S)

• Given solubility $$= x { g L}^{-1}$$
• Given molar mass $$= y{ g mol}^{-1}$$
• Molar solubility $$S = $$ $$\frac{x}{y}molL^{-1}$$
  

Step 2: Determine Anion Concentration and $$K_{sp}$$

The salt $$M_{3}A_{2}$$ dissociate into ions:

$$M_3A_2\longleftrightarrow\ 3M^{2+}\left(aq\right)+2A^{3-}\left(aq\right)$$

Cation Concentration $$[M^{2+}]=3S$$

Anion Concentration $$[A^{3-}]= 2S$$

The Solubility Product is:

$$\ K_{sp}=\left[M^{2+}\right]^3\left[A^{3-}\right]^2$$

$$\ K_{sp}=(3S^3)\times\ (2S^2)$$

$$\ K_{sp}=27S^3\times\ 4S^2=108S^5$$

Step 3: Calculate the Required Ratio

Ratio $$=\frac{[A^{3-}]}{K_{sp}}=\frac{2S}{108S^5}=\frac{1}{54S^4}$$

Substitute the value of S in equation:

Ratio $$=\frac{1}{54\left(\frac{x}{y}\right)^4}=\frac{y^4}{54x^4}=\frac{1}{54}\times\ \frac{y^4}{x^4}$$

Therefore, Option A is correct