If `100 mL` of acidified `2NH_(2)O_(2)` is allowed to react with `KMnO_(4)` solution till there is light tinge of purples colour, the volume of oxygen produced at `STP` is :

Options

A. `2.24 L`
B. `1.12 L`
C. `3.36 L`
D. `4.48 L`

Explanation

To solve the problem of finding the volume of oxygen produced at STP when 100 mL of acidified 2N hydrogen peroxide (H2O2) reacts with KMnO4, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Normality and Volume Strength Relationship**:
- We know that 1 normality of hydrogen peroxide produces 5.6 volume strength of O2.
- Therefore, for 2 normality (2N) of H2O2, the volume strength can be calculated as:
\[ \text{Volume Strength} = 5.6 \, \text{(for 1N)} \times 2 = 11.2 \, \text{volume strength} \]

2. **Calculate the Volume of O2 from H2O2**:
- The volume of O2 produced can be calculated using the formula:
\[ \text{Volume of O2} = \text{Volume of H2O2} \times \text{Volume Strength} \]
- Given that the volume of H2O2 is 100 mL, we substitute the values:
\[ \text{Volume of O2} = 100 \, \text{mL} \times 11.2 = 1120 \, \text{mL} \]

3. **Consider the Reaction with KMnO4**:
- The reaction with KMnO4 will consume the H2O2 and produce the same amount of O2. Therefore, the volume of O2 produced by KMnO4 is also 1120 mL.

4. **Total Volume of O2 Produced**:
- Since both H2O2 and KMnO4 produce the same volume of O2, we can add the volumes:
\[ \text{Total Volume of O2} = 1120 \, \text{mL} + 1120 \, \text{mL} = 2240 \, \text{mL} \]

5. **Convert to Liters**:
- To convert milliliters to liters, we divide by 1000:
\[ \text{Total Volume in Liters} = \frac{2240 \, \text{mL}}{1000} = 2.24 \, \text{L} \]

### Final Answer:
The volume of oxygen produced at STP is **2.24 liters**.