A `3.4g` sample of `H_(2)O_(2)` solution containing `x%H_(2)O` by mass requires `xmL` of a `KMnO_(4)` solution for complete oxidation under acidic conditions. The molarity of `KMnO_(4)` solution is `:`

Options

A. 1
B. 0.5
C. 0.4
D. 0.2

Answer

To solve the problem step by step, we will follow the procedure outlined in the video transcript.

### Step 1: Determine the mass of H2O2 in the solution
Given that the solution has a mass of 3.4 g and contains x% H2O2 by mass, we can calculate the mass of H2O2 in the solution.

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

### Step 2: Write the balanced chemical reaction
The reaction between KMnO4 and H2O2 under acidic conditions can be represented as follows:

\[ \text{KMnO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{Mn}^{2+} + \text{O}_2 + \text{H}_2\text{O} \]

### Step 3: Determine the n-factor for KMnO4 and H2O2
- For KMnO4, manganese changes from +7 to +2, which means it gains 5 electrons. Thus, the n-factor for KMnO4 is 5.
- For H2O2, the oxidation state of oxygen changes from -1 to 0. Since there are 2 oxygen atoms, it loses 2 electrons. Thus, the n-factor for H2O2 is 2.

### Step 4: Set up the equivalence equation
At complete oxidation, the equivalence of KMnO4 will equal the equivalence of H2O2:

\[ \text{Equivalence of KMnO}_4 = \text{Equivalence of H}_2\text{O}_2 \]

The equivalence can be calculated using the formula:

\[ \text{Equivalence} = \text{Molarity} \times \text{Volume (L)} \times n \]

### Step 5: Calculate the equivalence of H2O2
The number of moles of H2O2 can be calculated as follows:

\[ \text{Number of moles of } H_2O_2 = \frac{\text{Mass of } H_2O_2}{\text{Molar mass of } H_2O_2} = \frac{\frac{x}{100} \times 3.4}{34} \]

Thus, the equivalence of H2O2 is:

\[ \text{Equivalence of } H_2O_2 = \text{Number of moles} \times n_{\text{factor}} = \left(\frac{\frac{x}{100} \times 3.4}{34}\right) \times 2 \]

### Step 6: Calculate the equivalence of KMnO4
Let M be the molarity of the KMnO4 solution. The equivalence of KMnO4 is:

\[ \text{Equivalence of KMnO}_4 = M \times \frac{x}{1000} \times 5 \]

### Step 7: Set the equivalences equal to each other
Setting the equivalences equal gives:

\[ M \times \frac{x}{1000} \times 5 = \left(\frac{\frac{x}{100} \times 3.4}{34}\right) \times 2 \]

### Step 8: Solve for M
Now we can simplify and solve for M:

1. Cancel x from both sides (assuming x ≠ 0):
\[ M \times \frac{5}{1000} = \frac{3.4 \times 2}{34 \times 100} \]

2. Simplifying the right side:
\[ M \times \frac{5}{1000} = \frac{6.8}{3400} \]

3. Rearranging gives:
\[ M = \frac{6.8 \times 1000}{3400 \times 5} \]

4. Simplifying further:
\[ M = \frac{6800}{17000} = 0.4 \text{ M} \]

### Final Answer:
The molarity of the KMnO4 solution is **0.4 M**.

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