0.2 g if a sample of `H_2 O_2` required 10 ml of 1 N `KMnO_4` for titration in acidic medium. The percentage purity of `H_2 O_2` sample is:

Options

A. 0.25
B. 0.85
C. 0.65
D. 0.95

Answer

To solve the problem of determining the percentage purity of the H₂O₂ sample, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of H₂O₂ sample = 0.2 g
- Volume of KMnO₄ used = 10 mL
- Normality of KMnO₄ = 1 N

2. **Determine the N-factor of H₂O₂:**
- In H₂O₂, oxygen is in the -1 oxidation state. When it is oxidized to O₂, the oxidation state changes to 0.
- The change in oxidation state for 2 oxygen atoms is:
\[ \text{Change} = (0 - (-1)) \times 2 = 2 \]
- Therefore, the N-factor (valency factor) for H₂O₂ is 2.

3. **Determine the N-factor of KMnO₄:**
- In KMnO₄, manganese (Mn) is in the +7 oxidation state and is reduced to +2.
- The change in oxidation state is:
\[ \text{Change} = (7 - 2) = 5 \]
- Therefore, the N-factor for KMnO₄ is 5.

4. **Calculate the Milliequivalents of KMnO₄:**
- Milliequivalents of KMnO₄ can be calculated using the formula:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (in L)} \]
- Convert 10 mL to liters:
\[ 10 \text{ mL} = 0.01 \text{ L} \]
- Now calculate:
\[ \text{Milliequivalents of KMnO₄} = 1 \, \text{N} \times 0.01 \, \text{L} = 0.01 \, \text{equivalents} = 10 \, \text{milliequivalents} \]

5. **Set Up the Equation for H₂O₂:**
- Since the milliequivalents of H₂O₂ will equal the milliequivalents of KMnO₄:
\[ \text{Milliequivalents of H₂O₂} = \text{N-factor of H₂O₂} \times \text{Millimoles of H₂O₂} \]
- Let the millimoles of H₂O₂ be \( x \):
\[ 2x = 10 \quad \Rightarrow \quad x = \frac{10}{2} = 5 \, \text{millimoles} \]

6. **Calculate the Mass of H₂O₂:**
- The molar mass of H₂O₂ is:
\[ \text{Molar mass of H₂O₂} = 2 \times 16 + 2 = 34 \, \text{g/mol} \]
- The mass of H₂O₂ can be calculated as:
\[ \text{Mass of H₂O₂} = \text{Millimoles} \times \text{Molar mass} = 5 \times \frac{34}{1000} = 0.17 \, \text{g} \]

7. **Calculate the Percentage Purity of H₂O₂:**
- The percentage purity is given by:
\[ \text{Percentage Purity} = \left( \frac{\text{Mass of H₂O₂}}{\text{Mass of sample}} \right) \times 100 \]
- Substitute the values:
\[ \text{Percentage Purity} = \left( \frac{0.17}{0.2} \right) \times 100 = 85\% \]

### Final Answer:
The percentage purity of the H₂O₂ sample is **85%**.