`3.4 g` sample of `H_(2)O_(2)` solution containing `x% H_(2)O_(2)` by weight requires `x mL of a KMnO_(4)` solution for complete oxidation under acidic condition. The normality of `KMnO_(4)` solution is

Options

A. 1N
B. 2N
C. 3N
D. 0.5N

Explanation

To solve the problem step by step, we will follow the logical flow of calculations based on the information provided in the question.

### Step 1: Determine the amount of H₂O₂ in the solution
We are given that a 3.4 g sample of H₂O₂ solution contains x% H₂O₂ by weight.

To find the mass of H₂O₂ in the sample:
\[ \text{Mass of H₂O₂} = \frac{x}{100} \times 3.4 \text{ g} \]

### Step 2: Calculate the equivalent weight of H₂O₂
The molecular weight of H₂O₂ (Hydrogen Peroxide) is calculated as follows:
- Molecular weight = 2 (for H) + 32 (for O) = 34 g/mol

The n-factor for H₂O₂ in acidic conditions is 2 (since it can donate 2 electrons).

Using the formula for equivalent weight:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{34}{2} = 17 \text{ g/equiv} \]

### Step 3: Calculate the number of equivalents of H₂O₂
Using the mass of H₂O₂ calculated in Step 1:
\[ \text{Number of equivalents of H₂O₂} = \frac{\text{Mass of H₂O₂}}{\text{Equivalent weight}} = \frac{\frac{x}{100} \times 3.4}{17} \]

### Step 4: Relate the equivalents of H₂O₂ to KMnO₄
According to the question, x mL of KMnO₄ solution is required for the complete oxidation of H₂O₂.

Let \( N \) be the normality of the KMnO₄ solution. The number of equivalents of KMnO₄ is given by:
\[ \text{Number of equivalents of KMnO₄} = N \times \frac{x}{1000} \]

### Step 5: Set up the equation
Since the number of equivalents of H₂O₂ is equal to the number of equivalents of KMnO₄:
\[ \frac{\frac{x}{100} \times 3.4}{17} = N \times \frac{x}{1000} \]

### Step 6: Simplify the equation
We can cancel \( x \) from both sides (assuming \( x \neq 0 \)):
\[ \frac{3.4}{17} = N \times \frac{1}{1000} \]

### Step 7: Solve for N
Rearranging the equation gives:
\[ N = \frac{3.4 \times 1000}{17} \]
Calculating this:
\[ N = \frac{3400}{17} = 200 \]

### Step 8: Final calculation
Thus, the normality \( N \) of the KMnO₄ solution is:
\[ N = 2 \text{ N} \]

### Conclusion
The normality of the KMnO₄ solution is **2 N**.

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