20mL of `H_(2)O_(2)` after acidification with dilute `H_(2)SO_(4)` required 30mL of N/12 `KMnO_(4)` for complete oxidation. Calculate the percentage of `H_(2)O_(2)` in the solution. Equivalent mass of `H_(2)O_(2)=17`.

Explanation

To solve the problem step by step, we need to calculate the percentage of hydrogen peroxide (H₂O₂) in the solution based on the given data.

### Step 1: Determine the Normality of KMnO₄ Solution
The question states that 30 mL of N/12 KMnO₄ is used.

- Normality (N) of KMnO₄ = N/12 = 1/12 N
- Volume of KMnO₄ used = 30 mL = 0.030 L

### Step 2: Calculate the Equivalent of KMnO₄
The equivalent of KMnO₄ can be calculated using the formula:
\[ \text{Equivalent} = \text{Normality} \times \text{Volume (L)} \]
\[ \text{Equivalent of KMnO₄} = \frac{1}{12} \times 0.030 = \frac{0.030}{12} = 0.0025 \text{ equivalents} \]

### Step 3: Write the Reaction Equation
In acidic medium, KMnO₄ oxidizes H₂O₂. The balanced reaction is:
\[ \text{H₂O₂} + \text{KMnO₄} \rightarrow \text{Products} \]
From the reaction, we know that 1 equivalent of H₂O₂ reacts with 1 equivalent of KMnO₄.

### Step 4: Calculate the Equivalent of H₂O₂
Since 0.0025 equivalents of KMnO₄ are used, it means that 0.0025 equivalents of H₂O₂ are present in the solution.

### Step 5: Calculate the Mass of H₂O₂
Using the equivalent mass of H₂O₂ (given as 17 g/equiv), we can calculate the mass of H₂O₂:
\[ \text{Mass of H₂O₂} = \text{Equivalents} \times \text{Equivalent mass} \]
\[ \text{Mass of H₂O₂} = 0.0025 \times 17 = 0.0425 \text{ g} \]

### Step 6: Calculate the Concentration of H₂O₂ in the Solution
The volume of the H₂O₂ solution is 20 mL or 0.020 L. To find the concentration (in g/L):
\[ \text{Concentration (g/L)} = \frac{\text{Mass of H₂O₂}}{\text{Volume of solution (L)}} \]
\[ \text{Concentration (g/L)} = \frac{0.0425 \text{ g}}{0.020 \text{ L}} = 2.125 \text{ g/L} \]

### Step 7: Calculate the Percentage of H₂O₂ in the Solution
To find the percentage of H₂O₂ in the solution, we use the formula:
\[ \text{Percentage} = \left( \frac{\text{Concentration (g/L)}}{\text{Density of solution (g/mL)}} \right) \times 100 \]
Assuming the density of the solution is approximately 1 g/mL (which is a common assumption for dilute solutions):
\[ \text{Percentage} = \left( \frac{2.125 \text{ g/L}}{1000 \text{ g/L}} \right) \times 100 = 0.2125\% \]

### Final Answer
The percentage of H₂O₂ in the solution is **0.2125%**.

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