0.1M solution of KI reacts with excess of `H_(2)SO_(4)` and `KIO_(3)` solutions, according to equation
`5I^(-)+IO_(3)^(-)+6H^(+)to3I_(2)+3H_(2)O`, which of the following statements is /are correct:

Options

A. 400 ml of the KI solution react with 0.004 mole of `KIO_3`
B. 100 ml of the KI solution reacts with 0.006 mole of `H_2SO_4`
C. 0.5 litre of the KI solution produced 0.005 mole of `I_2`
D. Equivalent weight of `KIO_3` is equal to `(("Molecular Weight")/5)`

Explanation

To solve the problem, we will analyze the given reaction and evaluate each statement based on stoichiometry and chemical principles.

### Given Reaction:
\[ 5I^- + IO_3^- + 6H^+ \rightarrow 3I_2 + 3H_2O \]

### Step 1: Calculate moles of KI in 400 mL solution
- Given concentration of KI = 0.1 M
- Volume of KI solution = 400 mL = 0.4 L

**Moles of KI:**
\[ \text{Moles of KI} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.4 \, \text{L} = 0.04 \, \text{mol} \]

### Step 2: Determine moles of \( IO_3^- \) required
From the balanced equation, we see that:
- 5 moles of \( I^- \) react with 1 mole of \( IO_3^- \).

**Moles of \( IO_3^- \) required:**
\[ \text{Moles of } IO_3^- = \frac{1}{5} \times \text{Moles of KI} = \frac{1}{5} \times 0.04 = 0.008 \, \text{mol} \]

### Step 3: Evaluate Statement A
**Statement A:** "400 mL of KI solution reacts with 0.004 mole of \( IO_3^- \)"
- We found that 0.008 moles of \( IO_3^- \) are required, so this statement is **false**.

### Step 4: Calculate moles of \( H_2SO_4 \) required for 100 mL of KI solution
- Volume of KI solution = 100 mL = 0.1 L

**Moles of KI:**
\[ \text{Moles of KI} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

From the balanced equation:
- 5 moles of \( I^- \) react with 6 moles of \( H^+ \).
- Since 1 mole of \( H_2SO_4 \) provides 2 moles of \( H^+ \), we need:
\[ \text{Moles of } H_2SO_4 = \frac{6}{2} \times \frac{1}{5} \times \text{Moles of KI} = \frac{3}{5} \times 0.01 = 0.006 \, \text{mol} \]

### Step 5: Evaluate Statement B
**Statement B:** "100 mL of KI solution reacts with 0.006 mole of \( H_2SO_4 \)"
- We found that 0.006 moles of \( H_2SO_4 \) are indeed required, so this statement is **true**.

### Step 6: Calculate moles of \( I_2 \) produced from 0.5 L of KI solution
- Volume of KI solution = 0.5 L

**Moles of KI:**
\[ \text{Moles of KI} = 0.1 \, \text{mol/L} \times 0.5 \, \text{L} = 0.05 \, \text{mol} \]

From the balanced equation:
- 5 moles of \( I^- \) produce 3 moles of \( I_2 \).

**Moles of \( I_2 \) produced:**
\[ \text{Moles of } I_2 = \frac{3}{5} \times \text{Moles of KI} = \frac{3}{5} \times 0.05 = 0.03 \, \text{mol} \]

### Step 7: Evaluate Statement C
**Statement C:** "0.5 liters of the KI solution produces 0.05 moles of \( I_2 \)"
- We found that 0.03 moles of \( I_2 \) are produced, so this statement is **false**.

### Step 8: Determine equivalent weight of \( KIO_3 \)
- The oxidation state of I in \( KIO_3 \) is +5 (as calculated).
- The n-factor for \( KIO_3 \) is the change in oxidation state, which is 5 (from +5 to 0).

**Equivalent weight of \( KIO_3 \):**
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{\text{Molecular weight}}{5} \]

### Step 9: Evaluate Statement D
**Statement D:** "The equivalent weight of \( KIO_3 \) is equal to molecular weight divided by 5"
- This is **true** based on our calculation.

### Final Conclusion:
- **Correct Statements:** B and D
- **Incorrect Statements:** A and C