100mL of 1M `KMnO_(4)` oxidised 100mL of `H_(2)O_(2)` in acidic medium ( when `MnO_(4)^(-)` is reduced to `Mn^(2+)` ) , voluem of same `KMnO_(4)` required to oxidise 100mL of `H_(2)O_(2)` in basic medium ( when `MnO_(4)^(-)` is reduced to `MnO_(2)` ) will be `:`

Options

A. `( 100//3) ` mL
B. `( 500 //3) `mL
C. `( 300 //5) `mL
D. 100mL

Answer

To solve the problem, we need to determine the volume of `KMnO4` required to oxidize 100 mL of `H2O2` in a basic medium, given that 100 mL of 1 M `KMnO4` oxidizes 100 mL of `H2O2` in an acidic medium.

### Step-by-Step Solution:

1. **Determine the n-factor for `KMnO4` in acidic medium:**
- In acidic medium, `MnO4^(-)` is reduced to `Mn^(2+)`.
- The oxidation state of Mn in `MnO4^(-)` is +7, and in `Mn^(2+)` it is +2.
- Change in oxidation state = +7 to +2 = 5.
- Therefore, the n-factor for `KMnO4` in acidic medium is 5.

2. **Calculate the normality of `KMnO4` in acidic medium:**
- Normality (N) = Molarity (M) × n-factor.
- Given Molarity of `KMnO4` = 1 M.
- N = 1 M × 5 = 5 N.

3. **Calculate the milliequivalents of `KMnO4` used:**
- Volume of `KMnO4` = 100 mL.
- Milliequivalents of `KMnO4` = Normality × Volume (in L).
- Milliequivalents = 5 N × 0.1 L = 0.5 milliequivalents.

4. **Determine the n-factor for `KMnO4` in basic medium:**
- In basic medium, `MnO4^(-)` is reduced to `MnO2`.
- The oxidation state of Mn in `MnO4^(-)` is +7, and in `MnO2` it is +4.
- Change in oxidation state = +7 to +4 = 3.
- Therefore, the n-factor for `KMnO4` in basic medium is 3.

5. **Set up the equation for milliequivalents in basic medium:**
- The milliequivalents of `KMnO4` in basic medium must equal the milliequivalents of `H2O2`.
- Milliequivalents of `H2O2` = Milliequivalents of `KMnO4` = 0.5 milliequivalents.

6. **Calculate the normality of `KMnO4` in basic medium:**
- Let the volume of `KMnO4` required in basic medium be V mL.
- Using the formula: Milliequivalents = Normality × Volume (in L).
- Normality of `KMnO4` in basic medium = Molarity × n-factor = 1 M × 3 = 3 N.
- Therefore, 0.5 = 3 N × (V/1000).
- Rearranging gives: V = (0.5 × 1000) / 3 = 166.67 mL.

### Final Answer:
The volume of `KMnO4` required to oxidize 100 mL of `H2O2` in basic medium is approximately **166.67 mL**.