Which salt may show the same value of vant Hoff factor (i)as that of `K_(4)Fe(CN)_(6)`in very dilute solution state?

Options

A. `AI_(2)(SO_(4))_(3)`
B. `NaCI`
C. `AI(NO_(3))_(3)`
D. `Na_(2)SO_(4)`

Explanation

To determine which salt may show the same value of van't Hoff factor (i) as that of \( K_4Fe(CN)_6 \) in very dilute solution state, we need to analyze the dissociation of \( K_4Fe(CN)_6 \) and compare it with other salts.

### Step-by-Step Solution:

1. **Identify the Dissociation of \( K_4Fe(CN)_6 \)**:
- The compound \( K_4Fe(CN)_6 \) dissociates in solution to produce:
\[ K_4Fe(CN)_6 \rightarrow 4K^+ + Fe(CN)_6^{4-} \]
- This results in a total of 5 ions (4 potassium ions and 1 complex ion).

2. **Determine the van't Hoff factor (i)**:
- The van't Hoff factor \( i \) is defined as the number of particles the solute breaks into in solution.
- For \( K_4Fe(CN)_6 \), \( i = 5 \).

3. **Analyze the Given Options**:
- We need to find a salt that also dissociates into 5 ions.

**Option A: \( Al_2(SO_4)_3 \)**:
- Dissociation:
\[ Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-} \]
- Total ions = 2 + 3 = 5. Thus, \( i = 5 \).

**Option B: \( NaCl \)**:
- Dissociation:
\[ NaCl \rightarrow Na^+ + Cl^- \]
- Total ions = 1 + 1 = 2. Thus, \( i = 2 \).

**Option C: \( Al(NO_3)_3 \)**:
- Dissociation:
\[ Al(NO_3)_3 \rightarrow Al^{3+} + 3NO_3^{-} \]
- Total ions = 1 + 3 = 4. Thus, \( i = 4 \).

**Option D: \( Na_2SO_4 \)**:
- Dissociation:
\[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \]
- Total ions = 2 + 1 = 3. Thus, \( i = 3 \).

4. **Conclusion**:
- The only salt that has the same van't Hoff factor \( i = 5 \) as \( K_4Fe(CN)_6 \) is \( Al_2(SO_4)_3 \).
- Therefore, the correct answer is **Option A: \( Al_2(SO_4)_3 \)**.