`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 the \( H_2O_2 \) solution, given that it reacts with \( 10 \, \text{mL} \) of \( N/0.56 \) \( MnO_4^- \) solution in acidic medium.

### Step-by-Step Solution:

1. **Understand the Concept of Milliequivalents**:
The milliequivalents of \( H_2O_2 \) will be equal to the milliequivalents of \( MnO_4^- \). This can be expressed using the formula:
\[ n_1 \cdot V_1 = n_2 \cdot V_2 \]
where \( n_1 \) and \( n_2 \) are the normalities of \( H_2O_2 \) and \( MnO_4^- \) respectively, and \( V_1 \) and \( V_2 \) are their respective volumes.

2. **Assign Known Values**:
- Volume of \( H_2O_2 \) solution, \( V_1 = 10 \, \text{mL} \)
- Volume of \( MnO_4^- \) solution, \( V_2 = 10 \, \text{mL} \)
- Normality of \( MnO_4^- \) solution, \( n_2 = \frac{1}{0.56} \)

3. **Set Up the Equation**:
Since the normality of \( H_2O_2 \) is not given, we denote it as \( n_{H_2O_2} \). The equation becomes:
\[ n_{H_2O_2} \cdot 10 = \left(\frac{1}{0.56}\right) \cdot 10 \]

4. **Simplify the Equation**:
We can simplify the equation by canceling out the \( 10 \, \text{mL} \) from both sides:
\[ n_{H_2O_2} = \frac{1}{0.56} \]

5. **Relate Normality to Volume Strength**:
The volume strength \( x \) of \( H_2O_2 \) can be related to its normality. We know that:
\[ \text{Volume strength} = \frac{x}{5.6} \, \text{N} \]
Therefore, we can write:
\[ n_{H_2O_2} = \frac{x}{5.6} \]

6. **Equate the Two Normalities**:
Now we equate the two expressions for normality:
\[ \frac{x}{5.6} = \frac{1}{0.56} \]

7. **Solve for \( x \)**:
Cross-multiplying gives:
\[ x = 5.6 \cdot \frac{1}{0.56} \]
Simplifying this:
\[ x = 5.6 \cdot \frac{100}{56} = 10 \]

8. **Conclusion**:
Therefore, the volume strength \( x \) of the \( H_2O_2 \) solution is \( 10 \).

### Final Answer:
\( x = 10 \)