A solution of `H_(2)O_(2)` labelled as '20 V' was left open. Due to this some, `H_(2)O_(2)` decomposed and volume strength of the solution decreased . To determine the new volume strength of the `H_(2)O_(2)` solution, 10 mL of the solution was taken and it was diluted to 100 mL . 10 mL of this diluted solution was titrated against 25 mL of 0.0245 M `KMnO_(4)` solution under acidic condition. Calculate the volume strength of the `H_(2)O_(2)` solution .

Options

A. 15.00 V
B. 17.15 V
C. 20.00 V
D. 12.30 V

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a solution of hydrogen peroxide (H₂O₂) that was originally labeled as '20 V', which means it has a volume strength of 20. Due to decomposition, we need to find the new volume strength after some H₂O₂ has decomposed.

2. **Dilution of the Solution**:
- We take 10 mL of the decomposed H₂O₂ solution and dilute it to 100 mL.
- Let the normality of the original solution be \( x \) (before dilution).
- After dilution, the normality will be \( x' \).
- Using the dilution formula:
\[ n_1 V_1 = n_2 V_2 \]
where \( n_1 = x \), \( V_1 = 10 \) mL, \( n_2 = x' \), and \( V_2 = 100 \) mL, we can write:
\[ x \cdot 10 = x' \cdot 100 \]
Simplifying this gives:
\[ x' = \frac{x}{10} \]

3. **Titration with KMnO₄**:
- We titrate 10 mL of the diluted solution against 25 mL of 0.0245 M KMnO₄ solution under acidic conditions.
- The normality of KMnO₄ can be calculated using its molarity and n-factor. The n-factor for KMnO₄ (which changes from MnO₄⁻ to Mn²⁺) is 5.
- Therefore, the normality \( N \) of KMnO₄ is:
\[ N = \text{Molarity} \times \text{n-factor} = 0.0245 \times 5 = 0.1225 \, \text{N} \]

4. **Using the Titration Formula**:
- Now we apply the titration formula again:
\[ n_1 V_1 = n_2 V_2 \]
where \( n_1 = x' \), \( V_1 = 10 \) mL, \( n_2 = 0.1225 \) N, and \( V_2 = 25 \) mL.
- Plugging in the values:
\[ x' \cdot 10 = 0.1225 \cdot 25 \]
- This simplifies to:
\[ x' \cdot 10 = 3.0625 \]
Therefore:
\[ x' = 0.30625 \, \text{N} \]

5. **Finding the Original Normality**:
- From the dilution relationship, we know:
\[ x = 10 \cdot x' = 10 \cdot 0.30625 = 3.0625 \, \text{N} \]

6. **Calculating the Volume Strength**:
- The volume strength (V) of H₂O₂ is calculated using the formula:
\[ \text{Volume Strength} = \text{Normality} \times 5.6 \]
- Substituting the value of \( x \):
\[ V = 3.0625 \times 5.6 = 17.15 \, \text{V} \]

### Final Answer:
The volume strength of the H₂O₂ solution is **17.15 V**.