A sample of `H_(2)O_(2)` is x% by mass x ml of `KMnO_(4)` are required to oxidize one gram of this `H_(2)O_(2)` sample. Calculate the normality of `KMnO_(4)` solution.

Options

A. 0.25 N
B. 0.49 N
C. 0.59 N
D. 1.20 N

Answer

To calculate the normality of the `KMnO4` solution required to oxidize a sample of `H2O2`, we can follow these steps:

### Step 1: Determine the mass of `H2O2` in the sample
Given that the sample of `H2O2` is `x%` by mass, we can express the mass of `H2O2` in 1 gram of the sample:
- Mass of `H2O2` in 1 gram of sample = \( \frac{x}{100} \) grams.

### Step 2: Calculate the equivalent weight of `H2O2`
The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n factor}} \]
For `H2O2`:
- Molecular weight of `H2O2` = 34 g/mol.
- n factor of `H2O2` (for oxidation) = 2.

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

### Step 3: Calculate the equivalents of `H2O2`
Using the equivalent weight, we can find the equivalents of `H2O2` in the sample:
\[ \text{Equivalents of } H2O2 = \frac{\text{Given weight}}{\text{Equivalent weight}} = \frac{\frac{x}{100}}{17} = \frac{x}{1700} \text{ equivalents}. \]

### Step 4: Relate the equivalents of `H2O2` to `KMnO4`
According to the stoichiometry of the reaction, the equivalents of `H2O2` will be equal to the equivalents of `KMnO4` used for oxidation:
\[ \text{Equivalents of } KMnO4 = \frac{x}{1700}. \]

### Step 5: Calculate the normality of `KMnO4`
Normality (N) is defined as the number of equivalents per liter of solution. Given that `x mL` of `KMnO4` solution is used, we convert this to liters:
\[ \text{Volume of } KMnO4 = \frac{x}{1000} \text{ L}. \]
Now, we can express the normality of `KMnO4`:
\[ N = \frac{\text{Equivalents of } KMnO4}{\text{Volume of } KMnO4 \text{ in liters}} = \frac{\frac{x}{1700}}{\frac{x}{1000}}. \]
Simplifying this gives:
\[ N = \frac{x}{1700} \times \frac{1000}{x} = \frac{1000}{1700} = \frac{10}{17}. \]

### Step 6: Calculate the numerical value of normality
Calculating the numerical value:
\[ N \approx 0.588 \text{ N} \approx 0.59 \text{ N}. \]

Thus, the normality of the `KMnO4` solution is approximately **0.59 N**.

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