A bottle of `H_2O_2` is labelled as `10 vol H_2O_2. 112mL` of this solution of `H_2O_2` is titrated against `0.04M` acidified solution of `KMnO_4`. Calculate the volume of `KMnO_4` in terms of litre.

Explanation

To solve the problem step by step, we will follow the chemical reaction and the stoichiometry involved in the titration of hydrogen peroxide (H₂O₂) with acidified potassium permanganate (KMnO₄).

### Step 1: Understand the Reaction
The balanced chemical reaction between hydrogen peroxide and acidified KMnO₄ is:
\[ 5 \text{H}_2\text{O}_2 + 2 \text{KMnO}_4 + 3 \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2 \text{MnSO}_4 + 8 \text{H}_2\text{O} + 5 \text{O}_2 \]

From this equation, we can see that 2 moles of KMnO₄ react with 5 moles of H₂O₂.

### Step 2: Calculate the Normality of H₂O₂
The label on the bottle states that it is "10 vol H₂O₂". This means that 1 liter of this solution will release 10 liters of oxygen gas (O₂) at standard conditions.

Using the relationship between volume strength and normality:
- 1 volume strength of H₂O₂ corresponds to 1/5.6 normality.
- Therefore, 10 volume strength corresponds to:
\[ \text{Normality of H}_2\text{O}_2 = \frac{10}{5.6} \text{ N} \]

### Step 3: Calculate the Volume of H₂O₂ in Liters
The volume of H₂O₂ used in the titration is given as 112 mL. To convert this to liters:
\[ V_1 = \frac{112 \text{ mL}}{1000} = 0.112 \text{ L} \]

### Step 4: Set Up the Equation for Equivalents
Using the concept of equivalents, we know:
\[ \text{Equivalents of H}_2\text{O}_2 = \text{Equivalents of KMnO}_4 \]

The formula for equivalents is:
\[ \text{Equivalents} = \text{Normality} \times \text{Volume (L)} \]

For H₂O₂:
\[ \text{Equivalents of H}_2\text{O}_2 = \left(\frac{10}{5.6}\right) \times 0.112 \]

For KMnO₄, we have:
\[ \text{Equivalents of KMnO}_4 = N_2 \times V_2 \]
Where \( N_2 \) is the normality of KMnO₄ and \( V_2 \) is the volume in liters.

### Step 5: Calculate Normality of KMnO₄
The normality of KMnO₄ can be calculated using its molarity and n-factor:
- The molarity of KMnO₄ is given as 0.04 M.
- The n-factor for KMnO₄ in this reaction is 5 (as it gains 5 electrons).

Thus, the normality of KMnO₄ is:
\[ N_2 = 0.04 \times 5 = 0.2 \text{ N} \]

### Step 6: Set Up the Equation
Now we can set up the equation:
\[ \left(\frac{10}{5.6}\right) \times 0.112 = 0.2 \times V_2 \]

### Step 7: Solve for \( V_2 \)
Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{\left(\frac{10}{5.6}\right) \times 0.112}{0.2} \]

Calculating the right side:
1. Calculate \( \frac{10}{5.6} \):
\[ \frac{10}{5.6} \approx 1.7857 \]
2. Multiply by 0.112:
\[ 1.7857 \times 0.112 \approx 0.199 \]
3. Divide by 0.2:
\[ V_2 \approx \frac{0.199}{0.2} \approx 0.995 \text{ L} \]

### Final Answer
Thus, the volume of KMnO₄ used in the titration is approximately:
\[ V_2 \approx 1 \text{ L} \]