`A_(3)B_(2)` is a sparingly soluble salt with molar mass `M(gmol_(-))` and solubility `x` gm `litre_(-1)`, the ratio of the molar concentration of `B^(3-)` to the solubilty product of the salt is `:-`

Options

A. `108(x^(5))/(M^(5))`
B. `(1)/(108)(x^(5))/(M^(5))`
C. `(1)/(54)(M^(4))/(x^(4))`
D. `54(M^(4))/(x^(4))`

Explanation

To solve the problem, we need to determine the ratio of the molar concentration of \( B^{3-} \) to the solubility product \( K_{sp} \) of the salt \( A_3B_2 \).

### Step-by-Step Solution:

1. **Understanding the Dissolution of the Salt:**
The salt \( A_3B_2 \) dissociates in water as follows:
\[ A_3B_2 (s) \rightleftharpoons 3A^{2+} (aq) + 2B^{3-} (aq) \]
From this equation, we can see that for every 1 mole of \( A_3B_2 \) that dissolves, it produces 3 moles of \( A^{2+} \) and 2 moles of \( B^{3-} \).

2. **Defining Solubility:**
Let the solubility of \( A_3B_2 \) be \( x \) g/L. The molar mass of the salt is \( M \) g/mol. Therefore, the molar solubility \( S \) in mol/L can be calculated as:
\[ S = \frac{x}{M} \]

3. **Calculating Molar Concentrations:**
From the dissolution equation:
- The concentration of \( A^{2+} \) ions will be \( 3S \).
- The concentration of \( B^{3-} \) ions will be \( 2S \).

4. **Writing the Solubility Product Expression:**
The solubility product \( K_{sp} \) for the salt can be expressed as:
\[ K_{sp} = [A^{2+}]^3 [B^{3-}]^2 \]
Substituting the concentrations:
\[ K_{sp} = (3S)^3 (2S)^2 = 27S^3 \cdot 4S^2 = 108S^5 \]

5. **Finding the Ratio:**
We need to find the ratio of the molar concentration of \( B^{3-} \) to the solubility product \( K_{sp} \):
\[ \text{Molar concentration of } B^{3-} = 2S \]
Thus, the ratio is:
\[ \text{Ratio} = \frac{[B^{3-}]}{K_{sp}} = \frac{2S}{108S^5} = \frac{2}{108S^4} = \frac{1}{54S^4} \]

### Final Answer:
The ratio of the molar concentration of \( B^{3-} \) to the solubility product \( K_{sp} \) is:
\[ \frac{1}{54S^4} \]