20 " mL of " `H_(2)O_(2)` is reacted completely with acidified `K_(2)Cr_(2)O_(7)` solution 40 " mL of " `K_(2)Cr_(3)O_(7)` solution was required to oxidised the `H_(2)O_(2)` completely. Also, 2.0 " mL of " the same `K_(2)Cr_(2)O_(7)` solution required 5.0 " mL of " a 1.0 M `H_(2)C_(2)O_(4)` solution to reach equivalence point. Which of the following statements is/are correct?

Options

A. The `H_(2)O_(2)` solution is 5 M.
B. The volume strength of `H_(2)O_(2)` is 56V.
C. The volume strength of `H_(2)O_(2)` is 112V.
D. If 40 " mL of " `(5M)/(8H_(2)O_(2))` is further added to the 10 " mL of " 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 in the question and apply stoichiometric principles.

### Step 1: Understanding the Reaction
We have 20 mL of H₂O₂ reacting with K₂Cr₂O₇. The stoichiometry of the reaction indicates that H₂O₂ is oxidized by K₂Cr₂O₇, and we need to find the equivalents of both reactants.

### Step 2: Calculate the Milliequivalents of H₂O₂
The milliequivalents of H₂O₂ can be calculated using the formula:
\[ \text{Milliequivalents} = \text{Volume (mL)} \times \text{Normality (N)} \]

Let’s denote the normality of H₂O₂ as \( N_1 \):
\[ \text{Milliequivalents of H₂O₂} = 20 \, \text{mL} \times N_1 \]

### Step 3: Calculate the Milliequivalents of K₂Cr₂O₇
From the problem, we know that 40 mL of K₂Cr₂O₇ is required to oxidize H₂O₂ completely. Let’s denote the normality of K₂Cr₂O₇ as \( N_2 \):
\[ \text{Milliequivalents of K₂Cr₂O₇} = 40 \, \text{mL} \times N_2 \]

### Step 4: Set Up the Equation
Since the milliequivalents of H₂O₂ and K₂Cr₂O₇ are equal at the equivalence point:
\[ 20 \, \text{mL} \times N_1 = 40 \, \text{mL} \times N_2 \]
This simplifies to:
\[ N_1 = 2N_2 \]

### Step 5: Analyzing the Second Reaction
We know that 2.0 mL of the same K₂Cr₂O₇ solution reacts with 5.0 mL of a 1.0 M H₂C₂O₄ solution. The normality of H₂C₂O₄ is 1.0 M (since it is a diprotic acid, its normality is also 1.0 N).

### Step 6: Calculate Milliequivalents for H₂C₂O₄
The milliequivalents of H₂C₂O₄ can be calculated as:
\[ \text{Milliequivalents of H₂C₂O₄} = 5.0 \, \text{mL} \times 1.0 \, \text{N} = 5 \, \text{meq} \]

### Step 7: Relate K₂Cr₂O₇ and H₂C₂O₄
Let’s denote the normality of K₂Cr₂O₇ as \( N_2 \):
\[ 2 \, \text{mL} \times N_2 = 5 \, \text{meq} \]
This simplifies to:
\[ N_2 = \frac{5}{2} = 2.5 \, \text{N} \]

### Step 8: Calculate Normality of H₂O₂
Now substituting \( N_2 \) back into the equation from Step 4:
\[ N_1 = 2N_2 = 2 \times 2.5 = 5 \, \text{N} \]

### Step 9: Volume Strength of H₂O₂
The volume strength of H₂O₂ can be calculated using the relationship:
1 N of H₂O₂ occupies 5.6 L of O₂. Therefore:
\[ 5 \, \text{N} \rightarrow 5 \times 5.6 = 28 \, \text{L} \]

### Step 10: Conclusion
Based on the calculations, we can conclude the following:
- The normality of H₂O₂ is 5 N.
- The volume strength of H₂O₂ is 28 L.