`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 \( 0.01 \, M \) \( KMnO_4 \) required to oxidize \( 100 \, mL \) of \( H_2O_2 \) in basic medium, given that the same volume of \( H_2O_2 \) is oxidized by \( 100 \, mL \) of \( 0.01 \, M \) \( KMnO_4 \) in acidic medium.

### Step 1: Calculate the equivalents of \( H_2O_2 \) in acidic medium

In acidic medium, the reaction can be represented as:
\[ MnO_4^{-} + 8H^+ + 5e^{-} \rightarrow Mn^{2+} + 4H_2O \]
Here, the n-factor for \( KMnO_4 \) is 5 because it gains 5 electrons.

Using the formula for equivalents:
\[ \text{Equivalents of } KMnO_4 = \text{Normality} \times \text{Volume} \]
Since the volume is \( 100 \, mL \) (or \( 0.1 \, L \)) and the molarity is \( 0.01 \, M \):
\[ \text{Normality of } KMnO_4 = Molarity \times n-factor = 0.01 \times 5 = 0.05 \, N \]
Thus, the equivalents of \( KMnO_4 \) used are:
\[ 0.05 \, N \times 0.1 \, L = 0.005 \, equivalents \]

### Step 2: Equate the equivalents in basic medium

In basic medium, the reaction is:
\[ MnO_4^{-} + 4H_2O + 3e^{-} \rightarrow MnO_2 + 8OH^{-} \]
Here, the n-factor for \( KMnO_4 \) is 3 because it gains 3 electrons.

Let \( V \) be the volume of \( KMnO_4 \) in \( mL \). The equivalents in basic medium can be calculated as:
\[ \text{Equivalents of } KMnO_4 = \text{Normality} \times \text{Volume} \]
The normality in basic medium is:
\[ \text{Normality} = Molarity \times n-factor = 0.01 \times 3 = 0.03 \, N \]
Thus, the equivalents of \( KMnO_4 \) used are:
\[ 0.03 \, N \times \frac{V}{1000} \, L = \frac{0.03V}{1000} \, equivalents \]

### Step 3: Set up the equation

Since the equivalents of \( H_2O_2 \) are the same in both reactions, we can equate the two:
\[ 0.005 = \frac{0.03V}{1000} \]

### Step 4: Solve for \( V \)

Rearranging the equation gives:
\[ V = \frac{0.005 \times 1000}{0.03} = \frac{5}{0.03} = \frac{500}{3} \, mL \]

### Conclusion

Thus, the volume \( V \) of \( 0.01 \, M \) \( KMnO_4 \) required in basic medium is:
\[ V = \frac{500}{3} \, mL \]