What weight of `KMnO_(4)` will be required to prepare 250 mL of its `N/10` solution if eq.wt of `KMnO_(4) " is " 31.6` ?

Explanation

To find the weight of KMnO₄ required to prepare a 250 mL of N/10 (0.1 N) solution, we can use the normality formula:

\[ \text{Normality (N)} = \frac{\text{Weight (g)}}{\text{Equivalent Weight (g/equiv)} \times \text{Volume (L)}} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Normality (N) = 0.1 N (which is N/10)
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