A bottle of `H_(2)O_(2)` is labelled as 10 vol `H_(2)O_(2)`. 112 " mL of " this solution of `H_(2)O_(2)` is titrated against 0.04 M acidified solution of `KMnO_(4)` the volume of `KMnO_(4)` in litre is

Options

A. 1 L
B. 2 L
C. 3 L
D. 4 L

Explanation

To solve the problem step by step, we will follow the stoichiometric principles and calculations involved in the titration of hydrogen peroxide (H₂O₂) with potassium permanganate (KMnO₄).

### Step 1: Understand the meaning of "10 volume H₂O₂"
A "10 volume" solution of H₂O₂ means that 1 liter of this solution can produce 10 liters of oxygen gas (O₂) when decomposed.

### Step 2: Calculate the moles of O₂ produced
From the definition of "10 volume," we can deduce that:
- 1 L of H₂O₂ produces 10 L of O₂.
- At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L.

Thus, the moles of O₂ produced from 1 L of H₂O₂ is:
\[ \text{Moles of O₂} = \frac{10 \text{ L}}{22.4 \text{ L/mol}} = \frac{10}{22.4} \text{ mol} \]

### Step 3: Calculate the moles of H₂O₂ in 112 mL
Since we have 112 mL of H₂O₂ solution, we need to find the moles of H₂O₂ in this volume.

First, convert 112 mL to liters:
\[ 112 \text{ mL} = 0.112 \text{ L} \]

Now, calculate the moles of H₂O₂:
\[ \text{Moles of H₂O₂} = \text{Moles of O₂} \times 2 = \left(\frac{10}{22.4}\right) \times 2 = \frac{20}{22.4} \text{ mol} \]

### Step 4: Determine the n-factor for H₂O₂
The n-factor for H₂O₂ in this reaction is calculated based on the change in oxidation state:
- The oxidation state of oxygen in H₂O₂ is -1, and in O₂ it is 0.
- The change in oxidation state for one oxygen atom is 1.
- Since there are 2 oxygen atoms in H₂O₂, the n-factor is:
\[ \text{n-factor for H₂O₂} = 2 \times 1 = 2 \]

### Step 5: Calculate the equivalents of H₂O₂
Using the n-factor, we can calculate the equivalents of H₂O₂:
\[ \text{Equivalents of H₂O₂} = \text{n-factor} \times \text{moles of H₂O₂} \]
\[ = 2 \times \left(\frac{20}{22.4} \times 0.112\right) \]

### Step 6: Set up the equivalence with KMnO₄
The reaction between H₂O₂ and KMnO₄ is:
\[ \text{Equivalents of H₂O₂} = \text{Equivalents of KMnO₄} \]
The n-factor for KMnO₄ in acidic medium is 5. Therefore:
\[ \text{Equivalents of KMnO₄} = 5 \times \text{moles of KMnO₄} \]

Let the volume of KMnO₄ solution in liters be \( V \). Then:
\[ 5 \times 0.04 \times V = 2 \times \left(\frac{20}{22.4} \times 0.112\right) \]

### Step 7: Solve for V
Now we can solve for \( V \):
\[ V = \frac{2 \times \left(\frac{20}{22.4} \times 0.112\right)}{5 \times 0.04} \]

Calculating this gives:
\[ V = \frac{2 \times \left(\frac{20 \times 0.112}{22.4}\right)}{0.2} \]
\[ = \frac{2 \times \left(\frac{2.24}{22.4}\right)}{0.2} \]
\[ = \frac{4.48}{4.48} = 1 \text{ L} \]

### Final Answer
The volume of KMnO₄ used in the titration is **1 liter**.

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