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. 0.0602
B. 0.1204
C. 0.24
D. `0.030`

Answer

To find the molarity of the hydrogen peroxide (H₂O₂) solution that is completely oxidized by the potassium permanganate (KMnO₄) solution, we can follow these steps:

### Step 1: Calculate the moles of KMnO₄ used
We know the molarity (M) and volume (V) of the KMnO₄ solution. The formula to calculate moles (n) is:

\[ n = M \times V \]

Given:
- Molarity of KMnO₄ = 0.01652 M
- Volume of KMnO₄ = 36.44 mL = 0.03644 L (conversion from mL to L)

Now, substituting the values:

\[ n_{KMnO₄} = 0.01652 \, \text{mol/L} \times 0.03644 \, \text{L} \]

Calculating this gives:

\[ n_{KMnO₄} = 0.0006019 \, \text{mol} \]

### Step 2: Determine the moles of KMnO₄ that reacted
From the balanced chemical reaction between KMnO₄ and H₂O₂ in acidic media, we know that:

\[ 2 \, \text{KMnO₄} + 5 \, \text{H₂O₂} \rightarrow 2 \, \text{Mn}^{2+} + 5 \, \text{O} + 6 \, \text{H}_2\text{O} \]

This indicates that 2 moles of KMnO₄ react with 5 moles of H₂O₂. Therefore, the mole ratio is:

\[ \frac{5 \, \text{moles of H₂O₂}}{2 \, \text{moles of KMnO₄}} \]

Using the moles of KMnO₄ calculated:

\[ n_{H₂O₂} = n_{KMnO₄} \times \frac{5}{2} \]

Substituting the value:

\[ n_{H₂O₂} = 0.0006019 \, \text{mol} \times \frac{5}{2} \]

Calculating this gives:

\[ n_{H₂O₂} = 0.00150475 \, \text{mol} \]

### Step 3: Calculate the molarity of H₂O₂
Now we can find the molarity of the H₂O₂ solution using the formula:

\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \]

Given:
- Volume of H₂O₂ = 25 mL = 0.025 L

Substituting the values:

\[ \text{Molarity of H₂O₂} = \frac{0.00150475 \, \text{mol}}{0.025 \, \text{L}} \]

Calculating this gives:

\[ \text{Molarity of H₂O₂} = 0.06019 \, \text{M} \]

### Final Answer
Thus, the molarity of the H₂O₂ solution is approximately **0.0602 M**.

---