The purity of `H_2O_2` in a given sample is `85%.` Calculate the weight of impure sample of `H_2O_2` which requires `10mL` of `M//5 KMnO_4` solution in a titration in acidic medium

Options

A. `2 g`
B. `0.2 g`
C. `0.17 g`
D. `0.15 g`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the milliequivalents of KMnO4 used
We know that the volume of KMnO4 solution used is 10 mL and its molarity is M/5.

Using the formula for milliequivalents:
\[ \text{Milliequivalents} = \text{Volume (mL)} \times \text{Normality (N)} \]

Since the molarity (M) is given as M/5, we need to convert it to normality (N). For KMnO4, which has a valency of 5 in acidic medium, the normality is equal to the molarity multiplied by 5.

\[ \text{Normality} = \frac{M}{5} \times 5 = M \]

Thus, the milliequivalents of KMnO4 used is:
\[ \text{Milliequivalents of KMnO4} = 10 \, \text{mL} \times \frac{M}{5} = 2 \, \text{mEq} \]

### Step 2: Relate the milliequivalents of KMnO4 to H2O2
From the titration reaction, we know that 1 mole of KMnO4 reacts with 1 mole of H2O2. Therefore, the milliequivalents of H2O2 will also be equal to the milliequivalents of KMnO4.

\[ \text{Milliequivalents of H2O2} = 2 \, \text{mEq} \]

### Step 3: Calculate the weight of H2O2
To find the weight of H2O2, we use the formula:
\[ \text{Weight (g)} = \text{Milliequivalents} \times \frac{\text{Equivalent weight}}{1000} \]

The equivalent weight of H2O2 can be calculated as follows:
- Molar mass of H2O2 = 34 g/mol
- n-factor for H2O2 = 2 (as it can donate 2 electrons in the reaction)

Thus, the equivalent weight of H2O2 is:
\[ \text{Equivalent weight} = \frac{34 \, \text{g/mol}}{2} = 17 \, \text{g/equiv} \]

Now substituting the values:
\[ \text{Weight of H2O2} = 2 \, \text{mEq} \times \frac{17 \, \text{g/equiv}}{1000} = 0.034 \, \text{g} \]

### Step 4: Calculate the weight of the impure sample
Given that the purity of H2O2 is 85%, we can find the weight of the impure sample using the formula:
\[ \text{Weight of impure sample} = \frac{\text{Weight of H2O2}}{\text{Purity}} = \frac{0.034 \, \text{g}}{0.85} \]

Calculating this gives:
\[ \text{Weight of impure sample} = \frac{0.034}{0.85} \approx 0.04 \, \text{g} \]

### Final Answer
The weight of the impure sample of H2O2 required is approximately **0.04 grams**.

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