Which of the following will have same value of van't Hoff factor as that of `K_4[Fe(CN)_6]` ?

Options

A. `Al_2(SO_4)_3`
B. `AlCl_3`
C. `Al(NO_3)_3`
D. `Al(OH)_3`

Explanation

To determine which of the given compounds has the same van't Hoff factor (i) as that of \( K_4[Fe(CN)_6] \), we need to analyze how each compound dissociates into ions in solution. The van't Hoff factor (i) is defined as the number of particles (ions or molecules) into which a compound dissociates in solution.

### Step-by-Step Solution:

1. **Identify the dissociation of \( K_4[Fe(CN)_6] \)**:
- The compound \( K_4[Fe(CN)_6] \) dissociates as follows:
\[ K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \]
- This results in a total of:
- 4 potassium ions (\( K^+ \))
- 1 hexacyanoferrate ion (\( [Fe(CN)_6]^{4-} \))
- Therefore, the total number of ions produced is:
\[ 4 + 1 = 5 \]
- Thus, the van't Hoff factor \( i \) for \( K_4[Fe(CN)_6] \) is 5.

2. **Calculate the van't Hoff factor for \( Al_2(SO_4)_3 \)**:
- The dissociation is:
\[ Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-} \]
- Total ions produced:
\[ 2 + 3 = 5 \]
- Therefore, \( i = 5 \).

3. **Calculate the van't Hoff factor for \( AlCl_3 \)**:
- The dissociation is:
\[ AlCl_3 \rightarrow Al^{3+} + 3Cl^{-} \]
- Total ions produced:
\[ 1 + 3 = 4 \]
- Therefore, \( i = 4 \).

4. **Calculate the van't Hoff factor for \( Al(NO_3)_3 \)**:
- The dissociation is:
\[ Al(NO_3)_3 \rightarrow Al^{3+} + 3NO_3^{-} \]
- Total ions produced:
\[ 1 + 3 = 4 \]
- Therefore, \( i = 4 \).

5. **Calculate the van't Hoff factor for \( Al(OH)_3 \)**:
- The dissociation is:
\[ Al(OH)_3 \rightarrow Al^{3+} + 3OH^{-} \]
- Total ions produced:
\[ 1 + 3 = 4 \]
- Therefore, \( i = 4 \).

### Conclusion:
Among the options provided, only \( Al_2(SO_4)_3 \) has the same van't Hoff factor \( i = 5 \) as \( K_4[Fe(CN)_6] \). Therefore, the correct answer is:

**Answer: \( Al_2(SO_4)_3 \)**