0.2 g of a sample of `H_(2)O_(2)` required 10 mL of 1 N `KMnO_(4)` in a titration in the presence of `H_(2)SO_(4)` Purity of `H_(2)O_(2)` is :

Explanation

To find the purity of the hydrogen peroxide (H₂O₂) sample, we can follow these steps:

### Step 1: Determine the equivalent of KMnO₄ used in the titration.
- Given that 10 mL of 1 N KMnO₄ is used, we can calculate the equivalents of KMnO₄.

\[ \text{Equivalents of KMnO₄} = \text{Normality} \times \text{Volume (in L)} = 1 \, \text{N} \times \frac{10 \, \text{mL}}{1000 \, \text{mL/L}} = 0.01 \, \text{equivalents} \]

### Step 2: Set up the equivalence between H₂O₂ and KMnO₄.
- The reaction between H₂O₂ and KMnO₄ in acidic medium shows that 1 equivalent of H₂O₂ reacts with 1 equivalent of KMnO₄. Thus, the equivalents of H₂O₂ in the sample will also be 0.01 equivalents.

### Step 3: Calculate the mass of pure H₂O₂ in the sample.
- The formula for equivalents of H₂O₂ is given by:

\[ \text{Equivalents of H₂O₂} = \frac{\text{mass of H₂O₂ (g)}}{\text{molar mass of H₂O₂ (g/mol)} \times \text{valency factor}} \]

- The molar mass of H₂O₂ is 34 g/mol, and the valency factor is 2 (since H₂O₂ can donate 2 electrons). Therefore:

\[ 0.01 = \frac{x}{34 \times 2} \]

- Rearranging gives:

\[ x = 0.01 \times 34 \times 2 = 0.68 \, \text{g} \]

### Step 4: Calculate the purity of the H₂O₂ sample.
- The purity percentage can be calculated using the formula:

\[ \text{Purity (\%)} = \left( \frac{\text{mass of pure H₂O₂}}{\text{mass of sample}} \right) \times 100 \]

- Substituting the known values:

\[ \text{Purity (\%)} = \left( \frac{0.68}{0.2} \right) \times 100 = 85\% \]

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

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