`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. `2 N`
C. `3 N`
D. `0.5 N`

Explanation

To solve the problem, we need to determine the normality of the KMnO4 solution used to oxidize a sample of H2O2. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the weight of H2O2 in the sample
Given that the sample weighs `3.4 g` and contains `x%` H2O2 by weight, we can calculate the weight of H2O2 in the sample.

\[ \text{Weight of } H_2O_2 = \frac{3.4 \, \text{g} \times x}{100} \]

### Step 2: Calculate the equivalent weight of H2O2
The molecular weight of H2O2 (Hydrogen Peroxide) is `34 g/mol`. The n-factor for H2O2 in acidic conditions is `2` (since it can donate two electrons). Therefore, the equivalent weight can be calculated as:

\[ \text{Equivalent weight of } H_2O_2 = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{34}{2} = 17 \, \text{g/equiv} \]

### Step 3: Calculate the equivalent moles of H2O2
Using the weight of H2O2 calculated in Step 1 and the equivalent weight from Step 2, we can find the equivalent moles of H2O2:

\[ \text{Equivalent moles of } H_2O_2 = \frac{\text{Weight of } H_2O_2}{\text{Equivalent weight}} = \frac{\frac{3.4 \, x}{100}}{17} = \frac{3.4 \, x}{1700} = 2x \times 10^{-3} \, \text{equiv} \]

### Step 4: Set up the normality equation
We know that the normality (N) multiplied by the volume (V) gives the equivalent moles. For KMnO4, we denote its normality as \( N_1 \) and its volume as \( V_1 \) (which is \( x \, \text{mL} \) or \( \frac{x}{1000} \, \text{L} \)). The normality of H2O2 can be denoted as \( N_2 \) (which is \( 2 \, \text{equiv} \) from the equivalent moles calculated).

Using the equation:

\[ N_1 V_1 = N_2 V_2 \]

Substituting the known values:

\[ N_1 \left(\frac{x}{1000}\right) = 2x \times 10^{-3} \]

### Step 5: Solve for the normality of KMnO4
Rearranging the equation to solve for \( N_1 \):

\[ N_1 = \frac{2x \times 10^{-3}}{\frac{x}{1000}} = 2 \times 10^{-3} \times \frac{1000}{x} = 2 \, \text{N} \]

Thus, the normality of the KMnO4 solution is:

\[ \text{Normality of } KMnO_4 = 2 \, \text{N} \]

### Final Answer:
The normality of the KMnO4 solution is **2 N**.

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