If 36.44 ml of `"0.01652 M KMnO"^(4)` solution in acid media is required to completely oxidize 25 ml of a `H_(2)O_(2)` solution. What will be the molarity of `H_(2)O_(2)` solution?

Options

A. 0.602
B. 0.1204
C. 0.24
D. `0.030`

Answer

To find the molarity of the \( H_2O_2 \) solution, we can follow these steps:

### Step 1: Calculate the moles of \( KMnO_4 \)
We know the molarity (M) and volume (V) of the \( KMnO_4 \) solution. The formula for calculating moles (n) is:

\[ n = M \times V \]

Given:
- Molarity of \( KMnO_4 = 0.01652 \, M \)
- Volume of \( KMnO_4 = 36.44 \, ml = 0.03644 \, L \) (convert ml to L by dividing by 1000)

Now, substituting the values:

\[ n_{KMnO_4} = 0.01652 \, M \times 0.03644 \, L = 0.0006019 \, moles \approx 0.000602 \, moles \]

### Step 2: Determine the stoichiometry of the reaction
The balanced chemical reaction between \( KMnO_4 \) and \( H_2O_2 \) in acidic medium is:

\[ 2 \, MnO_4^- + 5 \, H_2O_2 + 6 \, H^+ \rightarrow 2 \, Mn^{2+} + 5 \, O_2 + 8 \, H_2O \]

From the reaction, we see that:
- 2 moles of \( MnO_4^- \) react with 5 moles of \( H_2O_2 \).

### Step 3: Calculate the moles of \( H_2O_2 \)
Using the stoichiometric ratio from the balanced equation:

\[ \frac{5 \, moles \, H_2O_2}{2 \, moles \, MnO_4^-} = \frac{x \, moles \, H_2O_2}{0.000602 \, moles \, MnO_4^-} \]

Cross-multiplying gives:

\[ x = 0.000602 \, moles \times \frac{5}{2} = 0.001505 \, moles \, H_2O_2 \]

### Step 4: Calculate the molarity of \( H_2O_2 \)
Now, we need to find the molarity of the \( H_2O_2 \) solution. Molarity (M) is defined as:

\[ M = \frac{n}{V} \]

Where:
- \( n = 0.001505 \, moles \)
- Volume of \( H_2O_2 = 25 \, ml = 0.025 \, L \)

Substituting the values:

\[ M_{H_2O_2} = \frac{0.001505 \, moles}{0.025 \, L} = 0.0602 \, M \]

### Final Answer
The molarity of the \( H_2O_2 \) solution is \( 0.0602 \, M \).

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