The volume of `10` volume `H_(2)O_(2)` solution that decolourises `200 mL` of `2 N KMnO_(4)` solution in acidic medium is

Options

A. 112 mL
B. 336 mL
C. 220 mL
D. 224 mL

Explanation

To solve the problem, we need to find the volume of a 10 volume H₂O₂ solution that can decolorize 200 mL of a 2 N KMnO₄ solution in an acidic medium. We will follow these steps:

### Step 1: Understand the Concept of Normality
Normality (N) is a measure of concentration equivalent to the number of equivalents of solute per liter of solution. For H₂O₂, we need to find its normality based on the given 10 volume concentration.

### Step 2: Calculate the Normality of H₂O₂
A 10 volume H₂O₂ solution means that 1 liter of H₂O₂ will produce 10 liters of O₂ at STP.

1. **Calculate moles of O₂ produced:**
- 10 liters of O₂ corresponds to \( \frac{10}{22.4} \) moles of O₂ (since 1 mole of gas occupies 22.4 L at STP).
- Therefore, moles of O₂ = \( \frac{10}{22.4} \approx 0.4464 \) moles.

2. **Relate moles of H₂O₂ to moles of O₂:**
- The reaction is: \( 2 H₂O₂ \rightarrow 2 H₂O + O₂ \)
- From the stoichiometry, 1 mole of H₂O₂ produces 0.5 moles of O₂.
- Therefore, moles of H₂O₂ = \( 0.4464 \times 2 \approx 0.8928 \) moles.

3. **Calculate the mass of H₂O₂:**
- Mass of H₂O₂ = moles × molar mass = \( 0.8928 \times 34 \approx 30.3 \) grams.

4. **Calculate the equivalent weight of H₂O₂:**
- Equivalent weight = Molar mass / n (where n = number of electrons transferred in the reaction).
- For H₂O₂, n = 2 (since it can donate 2 electrons).
- Equivalent weight = \( \frac{34}{2} = 17 \) grams/equiv.

5. **Calculate Normality of H₂O₂:**
- Normality (N) = \( \frac{\text{mass of solute (g)}}{\text{equivalent weight (g/equiv)} \times \text{volume (L)}} \)
- Assuming we consider 1 L of H₂O₂ solution,
- Normality = \( \frac{30.3}{17} \approx 1.78 \) N.

### Step 3: Use the Relation N₁V₁ = N₂V₂
Where:
- N₁ = Normality of H₂O₂ = 1.78 N
- V₁ = Volume of H₂O₂ (what we need to find)
- N₂ = Normality of KMnO₄ = 2 N
- V₂ = Volume of KMnO₄ = 200 mL

Using the formula:
\[ N₁V₁ = N₂V₂ \]
\[ 1.78 \times V₁ = 2 \times 200 \]

### Step 4: Solve for V₁
\[ V₁ = \frac{2 \times 200}{1.78} \]
\[ V₁ = \frac{400}{1.78} \approx 224.7 \text{ mL} \]

### Step 5: Round to the Nearest Option
The closest option to 224.7 mL is 224 mL.

### Final Answer
The volume of 10 volume H₂O₂ solution that decolorizes 200 mL of 2 N KMnO₄ solution is **224 mL**.

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