A 1 g sample of `H_(2)O_(2)` solution containing x % `H_(2)O_(2)` by mass requires x `cm^(3)` of a `KMnO_(4)` solution for complete oxidation under acidic conditions. Calculate the normality of `KMnO_(4)` solution for complete oxidation under acidic conditions. Calculate the normality of `KMnO_(4)` solution.

Options

A. 0.588 N
B. 0.294 N
C. 0.882 N
D. 0.735 N

Explanation

To solve the problem, we need to find the normality of the KMnO₄ solution used for the complete oxidation of a 1 g sample of H₂O₂ solution containing x% H₂O₂ by mass. Here’s the step-by-step solution:

### Step 1: Determine the mass of H₂O₂ in the solution
Given that the solution contains x% H₂O₂ by mass, the mass of H₂O₂ in the 1 g sample can be calculated as:
\[ \text{Mass of H₂O₂} = \frac{x}{100} \times 1 \text{ g} = \frac{x}{100} \text{ g} \]

### Step 2: Calculate the equivalent mass of H₂O₂
The equivalent mass of H₂O₂ can be calculated using its molar mass. The molar mass of H₂O₂ (Hydrogen Peroxide) is approximately 34 g/mol. Since H₂O₂ can release 2 electrons during oxidation, its equivalent mass is:
\[ \text{Equivalent mass of H₂O₂} = \frac{\text{Molar mass}}{\text{n}} = \frac{34 \text{ g/mol}}{1} = 34 \text{ g/equiv} \]

### Step 3: Calculate the milliequivalents of H₂O₂
The number of milliequivalents of H₂O₂ in the solution can be calculated as:
\[ \text{Milliequivalents of H₂O₂} = \frac{\text{Mass of H₂O₂}}{\text{Equivalent mass}} \times 1000 \]
Substituting the values we have:
\[ \text{Milliequivalents of H₂O₂} = \frac{\frac{x}{100}}{34} \times 1000 = \frac{10x}{34} = \frac{5x}{17} \text{ meq} \]

### Step 4: Relate the milliequivalents of KMnO₄ to those of H₂O₂
According to the problem, the milliequivalents of KMnO₄ used for the reaction is equal to the milliequivalents of H₂O₂:
\[ \text{Milliequivalents of KMnO₄} = \text{Milliequivalents of H₂O₂} \]
Let N be the normality of the KMnO₄ solution and the volume of KMnO₄ used is x cm³:
\[ N \times x = \frac{5x}{17} \]

### Step 5: Solve for the normality (N) of KMnO₄
We can simplify the equation:
\[ N \times x = \frac{5x}{17} \]
Dividing both sides by x (assuming x ≠ 0):
\[ N = \frac{5}{17} \text{ N} \]
To convert this to a decimal:
\[ N \approx 0.294 \text{ N} \]

### Step 6: Final Calculation
To express this in a more standard form, we can convert it to a fraction:
\[ N \approx 0.294 \text{ N} \text{ (or 0.294 equivalents per liter)} \]

### Conclusion
The normality of the KMnO₄ solution is approximately **0.294 N**.

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