`20mL` of `H_2O_2` is reacted completely with acidified `K_2Cr_2O_7` solution. `40mL` of `K_2Cr_2O_7` solution is required to oxidise the `H_2O_2` completely. Also, `2.0mL` of the same `K_2Cr_2O_7` solution is required to oxidise `5.0mL` of a `1.0MH_2C_2O_4` solution to reach equivalence point. Which of the following statements is`//`are correct?

Options

A. The `H_(2)O_(2)` solution is `5M`
B. The volume strength of `H_(2)O_(2)` is `56 V`
C. The volume strength of `H_(2)O_(2)` is `112 V`
D. If `40 mL` more `5M//8H_(2)O` is further added to the `10 mL` of the above `H_(2)O_(2)` solution, the volume strength of the resulting solution is changed to `16.8 V`

Explanation

To solve the problem step-by-step, we will analyze the information given and apply stoichiometry principles.

### Step 1: Determine the n-factor for each reactant
1. **For K2Cr2O7 (Potassium dichromate)**:
- The reduction half-reaction is:
\[ Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+} \]
- The n-factor is 6.

2. **For H2O2 (Hydrogen peroxide)**:
- The oxidation half-reaction is:
\[ H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O \]
- The n-factor is 2.

3. **For H2C2O4 (Oxalic acid)**:
- The oxidation half-reaction is:
\[ C_2O_4^{2-} \rightarrow 2CO_2 + 2e^- \]
- The n-factor is also 2.

### Step 2: Calculate the concentration of K2Cr2O7
- Given that 40 mL of K2Cr2O7 is required to oxidize 20 mL of H2O2:
\[ \text{Milliequivalents of } K_2Cr_2O_7 = n \times C \times V \]
- For K2Cr2O7:
\[ \text{Milliequivalents} = 6 \times C_{K_2Cr_2O_7} \times 40 \text{ mL} \]
- For H2O2:
\[ \text{Milliequivalents} = 2 \times C_{H_2O_2} \times 20 \text{ mL} \]
- Setting them equal gives:
\[ 6 \times C_{K_2Cr_2O_7} \times 40 = 2 \times C_{H_2O_2} \times 20 \]

### Step 3: Solve for the concentration of K2Cr2O7
- Rearranging the equation:
\[ C_{K_2Cr_2O_7} = \frac{2 \times C_{H_2O_2} \times 20}{6 \times 40} \]
- Simplifying:
\[ C_{K_2Cr_2O_7} = \frac{C_{H_2O_2}}{12} \]

### Step 4: Calculate the concentration of H2C2O4
- Given that 2 mL of K2Cr2O7 is required to oxidize 5 mL of 1.0 M H2C2O4:
\[ \text{Milliequivalents of } K_2Cr_2O_7 = 6 \times C_{K_2Cr_2O_7} \times 2 \]
\[ \text{Milliequivalents of } H_2C_2O_4 = 2 \times 1.0 \times 5 \]
- Setting them equal gives:
\[ 6 \times C_{K_2Cr_2O_7} \times 2 = 2 \times 1.0 \times 5 \]
- Solving for \( C_{K_2Cr_2O_7} \):
\[ C_{K_2Cr_2O_7} = \frac{5}{6} \]

### Step 5: Substitute back to find the concentration of H2O2
- From Step 3:
\[ C_{K_2Cr_2O_7} = \frac{C_{H_2O_2}}{12} \]
- Substituting \( C_{K_2Cr_2O_7} = \frac{5}{6} \):
\[ \frac{5}{6} = \frac{C_{H_2O_2}}{12} \]
- Solving for \( C_{H_2O_2} \):
\[ C_{H_2O_2} = \frac{5 \times 12}{6} = 10 \text{ M} \]

### Step 6: Calculate the volume strength of H2O2
- The volume strength is calculated as:
\[ \text{Volume strength} = 11.2 \times C_{H_2O_2} \]
- Thus:
\[ \text{Volume strength} = 11.2 \times 10 = 112 \text{ volume} \]

### Conclusion
- The correct statements based on the calculations are:
- A: Correct
- B: Correct
- D: Correct
- C: Incorrect