100 mL of 0.01 M `KMnO_(4)` oxidised 100 mL `H_(2)O_(2)` in acidic medium. The volume of same `KMnO_(4)` required in strong alkaline medium to oxidise 100 mL of same `H_(2)O_(2)` will be:

Options

A. `(100)/(3)mL`
B. `(500)/(3)mL`
C. `(300)/(5)mL`
D. none of these

Explanation

To solve the problem, we need to determine the volume of 0.01 M KMnO₄ required to oxidize 100 mL of H₂O₂ in a strong alkaline medium, given that the same amount of KMnO₄ oxidized the same volume of H₂O₂ in an acidic medium.

### Step-by-Step Solution:

1. **Determine the moles of KMnO₄ in acidic medium:**
- The concentration of KMnO₄ is given as 0.01 M and the volume is 100 mL.
- Convert the volume from mL to L:
\[ 100 \, \text{mL} = 0.1 \, \text{L} \]
- Calculate the moles of KMnO₄:
\[ \text{Moles of KMnO₄} = \text{Concentration} \times \text{Volume} = 0.01 \, \text{mol/L} \times 0.1 \, \text{L} = 0.001 \, \text{mol} \]

2. **Determine the equivalent weight of KMnO₄ in acidic medium:**
- In acidic medium, KMnO₄ is reduced from MnO₄⁻ (Mn in +7 oxidation state) to Mn²⁺ (Mn in +2 oxidation state).
- The change in oxidation state is 5, meaning it gains 5 electrons. Therefore, the n-factor (number of electrons exchanged) is 5.

3. **Calculate the equivalents of KMnO₄:**
- The number of equivalents of KMnO₄ used in the reaction is:
\[ \text{Equivalents} = \text{Moles} \times \text{n-factor} = 0.001 \, \text{mol} \times 5 = 0.005 \, \text{equivalents} \]

4. **Determine the equivalent weight of KMnO₄ in alkaline medium:**
- In alkaline medium, KMnO₄ is reduced from MnO₄⁻ (Mn in +7 oxidation state) to MnO₄²⁻ (Mn in +6 oxidation state).
- The n-factor in alkaline medium is 1 (as it gains only 1 electron).

5. **Set up the equation for KMnO₄ in alkaline medium:**
- Let \( V_2 \) be the volume of KMnO₄ required in alkaline medium.
- The equivalents of KMnO₄ in alkaline medium can be expressed as:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume} \times \text{n-factor} \]
- Thus, we have:
\[ 0.01 \, \text{mol/L} \times V_2 \times 1 = 0.005 \, \text{equivalents} \]

6. **Solve for \( V_2 \):**
- Rearranging the equation gives:
\[ V_2 = \frac{0.005 \, \text{equivalents}}{0.01 \, \text{mol/L}} = 0.5 \, \text{L} = 500 \, \text{mL} \]

### Final Answer:
The volume of 0.01 M KMnO₄ required in strong alkaline medium to oxidize 100 mL of the same H₂O₂ is **500 mL**.