To find the percentage purity of the hydrogen peroxide (H₂O₂) sample, we will follow these steps:
### Step 1: Write the balanced redox reaction
The reaction between hydrogen peroxide (H₂O₂) and potassium permanganate (KMnO₄) in acidic medium can be represented as follows:
\[ 2 \text{MnO}_4^- + 5 \text{H}_2\text{O}_2 + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 5 \text{O}_2 + 8 \text{H}_2\text{O} \]
### Step 2: Calculate the milliequivalents of KMnO₄ used
Given that 10 ml of 1 N KMnO₄ is used, we can calculate the milliequivalents:
\[
\text{Milliequivalents of KMnO}_4 = \text{Normality} \times \text{Volume in L}
\]
\[
= 1 \, \text{N} \times 0.010 \, \text{L} = 0.010 \, \text{equivalents} = 10 \, \text{milliequivalents}
\]
### Step 3: Determine the moles of KMnO₄
Since 1 equivalent of KMnO₄ reacts with 5 equivalents of H₂O₂, we can find the moles of KMnO₄:
\[
\text{Moles of KMnO}_4 = \frac{\text{Milliequivalents}}{n \text{ factor}} = \frac{10 \, \text{meq}}{5} = 2 \, \text{mmoles}
\]
### Step 4: Calculate the moles of H₂O₂ reacted
From the balanced equation, we see that 2 moles of KMnO₄ react with 5 moles of H₂O₂. Therefore, for 2 mmoles of KMnO₄, the moles of H₂O₂ will be:
\[
\text{Moles of H}_2\text{O}_2 = 2 \, \text{mmoles KMnO}_4 \times \frac{5 \, \text{mmoles H}_2\text{O}_2}{2 \, \text{mmoles KMnO}_4} = 5 \, \text{mmoles H}_2\text{O}_2
\]
### Step 5: Calculate the mass of H₂O₂
The molar mass of H₂O₂ is 34 g/mol. Therefore, the mass of H₂O₂ in the sample is:
\[
\text{Mass of H}_2\text{O}_2 = \text{Moles} \times \text{Molar Mass} = 5 \, \text{mmoles} \times \frac{34 \, \text{g}}{1000 \, \text{mmoles}} = 0.17 \, \text{g}
\]
### Step 6: Calculate the percentage purity
The percentage purity of the H₂O₂ sample can be calculated using the formula:
\[
\text{Percentage Purity} = \left( \frac{\text{Mass of H}_2\text{O}_2 \text{ in sample}}{\text{Total mass of sample}} \right) \times 100
\]
\[
= \left( \frac{0.17 \, \text{g}}{0.2 \, \text{g}} \right) \times 100 = 85\%
\]
### Final Answer
The percentage purity of the H₂O₂ sample is **85%**.
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