The `K_(sp)`of `K_(sp)` & `M(OH)_(3)` is `4xx10^(-21)`and `9xx10^(-24)`respectively. The electrolyte that precipitate first on adding `Zn(OH)_(2)` slowly in a solution of `0.1 M M^(2+)` & `0.1 M M^(3+)` `:-`

Options

A. `M(OH)_(2)`
B. `M(OH)_(3)`
C. Both together
D. Can't predict

Explanation

To solve the problem, we need to determine which hydroxide salt, \( M(OH)_2 \) or \( M(OH)_3 \), will precipitate first when adding \( Zn(OH)_2 \) to a solution containing \( 0.1 \, M \, M^{2+} \) and \( 0.1 \, M \, M^{3+} \). We will use the solubility product constants (\( K_{sp} \)) provided for both salts to find the concentration of hydroxide ions (\( OH^- \)) required for precipitation.

### Step-by-Step Solution:

1. **Identify the \( K_{sp} \) Values:**
- For \( M(OH)_2 \): \( K_{sp} = 4 \times 10^{-21} \)
- For \( M(OH)_3 \): \( K_{sp} = 9 \times 10^{-24} \)

2. **Write the Dissociation Equations:**
- For \( M(OH)_2 \):
\[ M(OH)_2 \rightleftharpoons M^{2+} + 2OH^- \]
The expression for \( K_{sp} \) is:
\[ K_{sp} = [M^{2+}][OH^-]^2 \]
- For \( M(OH)_3 \):
\[ M(OH)_3 \rightleftharpoons M^{3+} + 3OH^- \]
The expression for \( K_{sp} \) is:
\[ K_{sp} = [M^{3+}][OH^-]^3 \]

3. **Calculate \( [OH^-] \) for \( M(OH)_2 \):**
- Given \( [M^{2+}] = 0.1 \, M \):
\[ K_{sp} = [0.1][OH^-]^2 = 4 \times 10^{-21} \]
Rearranging gives:
\[ [OH^-]^2 = \frac{4 \times 10^{-21}}{0.1} = 4 \times 10^{-20} \]
Taking the square root:
\[ [OH^-] = \sqrt{4 \times 10^{-20}} = 2 \times 10^{-10} \, M \]

4. **Calculate \( [OH^-] \) for \( M(OH)_3 \):**
- Given \( [M^{3+}] = 0.1 \, M \):
\[ K_{sp} = [0.1][OH^-]^3 = 9 \times 10^{-24} \]
Rearranging gives:
\[ [OH^-]^3 = \frac{9 \times 10^{-24}}{0.1} = 9 \times 10^{-23} \]
Taking the cube root:
\[ [OH^-] = \sqrt[3]{9 \times 10^{-23}} \approx 4.48 \times 10^{-8} \, M \]

5. **Determine Which Precipitates First:**
- The \( OH^- \) concentration required for \( M(OH)_2 \) to precipitate is \( 2 \times 10^{-10} \, M \).
- The \( OH^- \) concentration required for \( M(OH)_3 \) to precipitate is approximately \( 4.48 \times 10^{-8} \, M \).
- Since \( 2 \times 10^{-10} \, M \) (for \( M(OH)_2 \)) is less than \( 4.48 \times 10^{-8} \, M \) (for \( M(OH)_3 \)), \( M(OH)_2 \) will precipitate first.

### Conclusion:
The electrolyte that precipitates first upon adding \( Zn(OH)_2 \) slowly in the solution is \( M(OH)_2 \).