The number of moles of `H_(2)O_(2)` required to completely react with 400 ml of 0.5 N `KMnO_(4)` in acidic medium are

Options

A. 0.1
B. 0.2
C. `1.0`
D. 0.5

Answer

To solve the problem of determining the number of moles of \( H_2O_2 \) required to completely react with 400 mL of 0.5 N \( KMnO_4 \) in acidic medium, we can follow these steps:

### Step 1: Understand the Reaction
In acidic medium, \( KMnO_4 \) acts as an oxidizing agent and reacts with \( H_2O_2 \). The n-factor (number of electrons transferred per molecule) for \( KMnO_4 \) in acidic medium is 5, and for \( H_2O_2 \), it is 2.

### Step 2: Calculate the Equivalents of \( KMnO_4 \)
First, we need to calculate the equivalents of \( KMnO_4 \) in the solution:
\[ \text{Equivalents of } KMnO_4 = \text{Normality} \times \text{Volume (in L)} \]
Given:
- Normality of \( KMnO_4 = 0.5 \, N \)
- Volume = 400 mL = 0.4 L

Calculating the equivalents:
\[ \text{Equivalents of } KMnO_4 = 0.5 \, N \times 0.4 \, L = 0.2 \, \text{equivalents} \]

### Step 3: Relate Equivalents of \( H_2O_2 \) to \( KMnO_4 \)
Since the reaction between \( KMnO_4 \) and \( H_2O_2 \) is a redox reaction, we can set the equivalents of \( H_2O_2 \) equal to the equivalents of \( KMnO_4 \):
\[ \text{Equivalents of } H_2O_2 = \text{Equivalents of } KMnO_4 \]
Thus:
\[ \text{Equivalents of } H_2O_2 = 0.2 \, \text{equivalents} \]

### Step 4: Calculate the Moles of \( H_2O_2 \)
Now we can find the number of moles of \( H_2O_2 \) using its n-factor:
\[ \text{Equivalents} = \text{Moles} \times \text{n-factor} \]
For \( H_2O_2 \), the n-factor is 2:
\[ 0.2 = \text{Moles of } H_2O_2 \times 2 \]
Solving for moles:
\[ \text{Moles of } H_2O_2 = \frac{0.2}{2} = 0.1 \, \text{moles} \]

### Conclusion
The number of moles of \( H_2O_2 \) required to completely react with 400 mL of 0.5 N \( KMnO_4 \) in acidic medium is **0.1 moles**.

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