For the electrolyte `A_(2)B" if "K_("sp")` is solubility product then its solubling ‘S’ M is `[K_("sp")]^(1//2) div 4`.

Answer

To solve the problem, we need to determine the relationship between the solubility product \( K_{sp} \) and the solubility \( S \) of the electrolyte \( A_2B \).

### Step-by-Step Solution:

1. **Dissociation of the Electrolyte**:
The electrolyte \( A_2B \) dissociates in solution as follows:
\[ A_2B \rightleftharpoons 2A^+ + B^{2-} \]
This means that for every 1 mole of \( A_2B \) that dissolves, it produces 2 moles of \( A^+ \) ions and 1 mole of \( B^{2-} \) ions.

2. **Define Solubility**:
Let the solubility of \( A_2B \) be \( S \) M. Therefore, when \( A_2B \) dissolves:
- The concentration of \( A^+ \) ions will be \( 2S \) (since 2 moles of \( A^+ \) are produced).
- The concentration of \( B^{2-} \) ions will be \( S \) (since 1 mole of \( B^{2-} \) is produced).

3. **Expression for \( K_{sp} \)**:
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [A^+]^2 [B^{2-}] \]
Substituting the concentrations from the previous step:
\[ K_{sp} = (2S)^2 (S) = 4S^2 \cdot S = 4S^3 \]

4. **Finding \( K_{sp}^{1/2} \)**:
Now, we need to find \( K_{sp}^{1/2} \):
\[ K_{sp}^{1/2} = (4S^3)^{1/2} = 2S^{3/2} \]

5. **Dividing by 4**:
Finally, we need to divide \( K_{sp}^{1/2} \) by 4:
\[ \frac{K_{sp}^{1/2}}{4} = \frac{2S^{3/2}}{4} = \frac{S^{3/2}}{2} \]

### Final Answer:
The final expression for the solubility \( S \) in terms of \( K_{sp} \) is:
\[ \frac{S^{3/2}}{2} \]