`100mL` of `H_2O_2` is oxidised by `100mL` of `0.01M KMnO_4` in acidic medium `(MnO_4^(ɵ)` reduced to `Mn^(2+)`). `100mL` of the same `H_2O_2` is oxidised by `VmL` of `0.01M KMnO_4` in basic medium. Hence `V` is

Options

A. 500
B. 100
C. `(100)/(3)`
D. `(500)/(3)`

Explanation

To solve the problem, we need to determine the volume \( V \) of \( 0.01M \, KMnO_4 \) required to oxidize \( 100mL \) of \( H_2O_2 \) in basic medium, given that the same volume of \( H_2O_2 \) is oxidized by \( 100mL \) of \( 0.01M \, KMnO_4 \) in acidic medium.

### Step-by-Step Solution:

1. **Determine the milliequivalents of KMnO4 in acidic medium**:
- The reaction of \( KMnO_4 \) in acidic medium involves the reduction of \( MnO_4^- \) to \( Mn^{2+} \).
- The n-factor for this reaction is 5 (since \( Mn \) goes from +7 to +2, which involves the transfer of 5 electrons).
- Calculate the milliequivalents of \( KMnO_4 \):
\[ \text{Milliequivalents of } KMnO_4 = \text{Normality} \times \text{Volume} = 0.01 \, \text{mol/L} \times 100 \, \text{mL} = 1 \, \text{meq} \]

2. **Set up the equivalence in acidic medium**:
- Since \( H_2O_2 \) is oxidized by \( KMnO_4 \), we can equate the milliequivalents of \( KMnO_4 \) to that of \( H_2O_2 \):
\[ \text{Milliequivalents of } H_2O_2 = 1 \, \text{meq} \]

3. **Determine the milliequivalents of KMnO4 in basic medium**:
- In basic medium, \( KMnO_4 \) is reduced to \( MnO_4^{2-} \).
- The n-factor for this reaction is 1 (since \( Mn \) goes from +7 to +6, which involves the transfer of 1 electron).
- The milliequivalents of \( KMnO_4 \) in basic medium can be calculated as:
\[ \text{Milliequivalents of } KMnO_4 = \text{Normality} \times \text{Volume} = 0.01 \, \text{mol/L} \times V \, \text{mL} \]

4. **Set up the equivalence in basic medium**:
- Equate the milliequivalents of \( KMnO_4 \) in basic medium to that of \( H_2O_2 \):
\[ 0.01 \times V = 1 \, \text{meq} \]

5. **Solve for \( V \)**:
- Rearranging the equation gives:
\[ V = \frac{1}{0.01} = 100 \, \text{mL} \]

6. **Conclusion**:
- The volume \( V \) of \( 0.01M \, KMnO_4 \) required in basic medium to oxidize \( 100mL \) of \( H_2O_2 \) is \( 100 \, \text{mL} \).

### Final Answer:
The value of \( V \) is \( 500 \, \text{mL} \).