Which of the following statements is/are correct :

Options

A. Equivalent mass of `Br_3O_8` in the reaction is 23.
`Br_3 O_8+I^(-)+H^+toBr_2+I_2+H_2O`
B. In case of oxalic acid `(H_2C_2O_4)`, the average oxidation number and individual oxidation number of Carbon have the same value.
C. In case of iodimetric titration, KI is made to react with an oxidising agent and the liberated `I_2` gas is made to titration with hypo solution.
D. Equivalent mass of `Cu_3P` in the given reaction is `M//11` (where M=mol. Mass)
`Cu_3P+KMnO_4+H^+toCu^(2+)+H_2PO_4^(-)+K^+ + Mn^(2+)`

Explanation

To determine which of the statements provided in the question are correct, we will analyze each statement step by step.

### Step 1: Evaluate the first statement about Br3O8
1. **Determine the oxidation state of Br in Br3O8**:
- The formula for calculating oxidation state is:
\[ 3x + 8(-2) = 0 \]
where \(x\) is the oxidation state of bromine (Br).
- Solving for \(x\):
\[ 3x - 16 = 0 \implies 3x = 16 \implies x = \frac{16}{3} \]

2. **Calculate the change in oxidation state**:
- The change in oxidation state for one Br atom going from \(\frac{16}{3}\) to 0 is:
\[ \Delta = 0 - \frac{16}{3} = -\frac{16}{3} \]
- For three Br atoms, the total change is:
\[ 3 \times \left(-\frac{16}{3}\right) = -16 \]

3. **Calculate the equivalent mass of Br3O8**:
- The equivalent mass is given by:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{n \text{ factor}} \]
- The molar mass of Br3O8 is calculated as follows:
- Molar mass of Br = 80 g/mol (approximately)
- Molar mass of O = 16 g/mol
- Therefore, molar mass of Br3O8 = \(3 \times 80 + 8 \times 16 = 240 + 128 = 368 \text{ g/mol}\)
- The n factor is 16 (from the total change in oxidation state).
- Thus, the equivalent mass is:
\[ \text{Equivalent mass} = \frac{368}{16} = 23 \]
- **Conclusion**: The first statement is correct.

### Step 2: Evaluate the second statement about oxalic acid
1. **Determine the oxidation state of carbon in oxalic acid (C2H2O4)**:
- The oxidation state of carbon can be calculated as follows:
\[ 2x + 2(+1) + 4(-2) = 0 \]
- Solving for \(x\):
\[ 2x + 2 - 8 = 0 \implies 2x = 6 \implies x = +3 \]
- In oxalic acid, there are two carbon atoms, and both have the same oxidation state of +3.

2. **Conclusion**: The average oxidation number of carbon is the same as the individual oxidation number, hence the second statement is correct.

### Step 3: Evaluate the third statement about iodometric titration
1. **Understanding the iodometric titration process**:
- In iodometric titration, potassium iodide (KI) reacts with an oxidizing agent to liberate iodine (I2).
- The liberated iodine is then titrated with sodium thiosulfate (Na2S2O3), which reduces iodine back to iodide ions (I-).

2. **Balanced reaction**:
- The reaction can be represented as:
\[ 2Na2S2O3 + I2 \rightarrow 2NaI + Na2S4O6 \]
- This indicates that iodine reacts with thiosulfate in a stoichiometric manner.

3. **Conclusion**: The third statement is correct as it accurately describes the process of iodometric titration.

### Step 4: Evaluate the fourth statement about Cu3P
1. **Determine the oxidation states in Cu3P**:
- The oxidation state of phosphorus (P) in Cu3P is -3.
- For copper (Cu), let the oxidation state be \(x\):
\[ 3x - 3 = 0 \implies 3x = 3 \implies x = +1 \]

2. **Identify the changes in oxidation states**:
- Copper changes from +1 to +2, contributing a change of +1 for each Cu atom:
\[ \text{Total change for Cu} = 3 \times 1 = 3 \]
- Phosphorus changes from -3 to +5, contributing a change of:
\[ 5 - (-3) = 8 \]

3. **Calculate the total n factor**:
- The total n factor is:
\[ n = 3 + 8 = 11 \]

4. **Calculate the equivalent mass of Cu3P**:
- The equivalent mass is given by:
\[ \text{Equivalent mass} = \frac{M}{n} = \frac{M}{11} \]
- Therefore, the fourth statement is also correct.

### Conclusion
All statements provided in the question are correct.