`10mL` of `H_2O_2` solution (volume strength `= x`) requires `10mL` of `N//0.56 MnO_4^(ɵ)` solution in acidic medium. Hence`x` is

Options

A. `0.56`
B. `5.6`
C. `0.1`
D. `10`

Explanation

To solve the problem, we need to find the volume strength \( x \) of a \( 10 \, \text{mL} \) solution of \( H_2O_2 \) that reacts with \( 10 \, \text{mL} \) of \( N/0.56 \, \text{MnO}_4^- \) solution in acidic medium.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
In acidic medium, the reaction between \( H_2O_2 \) and \( MnO_4^- \) can be represented as:
\[ H_2O_2 + MnO_4^- \rightarrow \text{products} \]
In this reaction, the number of equivalents of \( H_2O_2 \) will be equal to the number of equivalents of \( MnO_4^- \).

2. **Calculate Milliequivalents**:
The milliequivalents (mEq) of a solution can be calculated using the formula:
\[ \text{Milliequivalents} = N \times V \]
where \( N \) is the normality and \( V \) is the volume in liters.

3. **Setting Up the Equation**:
Let \( N_1 \) be the normality of \( H_2O_2 \) and \( N_2 \) be the normality of \( MnO_4^- \). Given that \( N_2 = \frac{1}{0.56} \) and both volumes are \( 10 \, \text{mL} \):
\[ N_1 \times 10 = N_2 \times 10 \]
Thus, we can simplify to:
\[ N_1 = N_2 \]

4. **Substituting Values**:
Substitute \( N_2 \):
\[ N_1 = \frac{1}{0.56} \]

5. **Finding Volume Strength**:
The volume strength \( x \) of \( H_2O_2 \) is related to its normality by the formula:
\[ x = 5.6 \times N_1 \]
Substituting \( N_1 \):
\[ x = 5.6 \times \frac{1}{0.56} \]

6. **Calculating \( x \)**:
\[ x = 5.6 \times \frac{1}{0.56} = 10 \]

### Conclusion:
Thus, the volume strength \( x \) of the \( H_2O_2 \) solution is \( 10 \).