The substance having the same value of van't Hoff factor as that of \(K_4[Fe(CN)_6]\) is:

\(Al_2(SO_4)_3\)

The correct answer is option 4. \(Al_2(SO_4)_3\).

The van't Hoff factor (\(i\)) represents the number of particles a compound dissociates into when dissolved in water. It's calculated as:

\(i = \frac{\text{Number of particles after dissociation}}{\text{Number of formula units initially dissolved}}\)

For example, if a compound dissociates completely into two ions, its van't Hoff factor would be \(2\).

Calculate the van't Hoff factor for \(K_4[Fe(CN)_6]\)

The compound \(K_4[Fe(CN)_6]\) dissociates in water as follows:

\(K_4[Fe(CN)_6] \longrightarrow 4K^+ + [Fe(CN)_6]^{4-}\)

This equation shows that one formula unit of \(K_4[Fe(CN)_6]\) produces:

4 potassium ions (\(K^+\))

1 hexacyanoferrate ion (\([Fe(CN)_6]^{4-}\))

Thus, the total number of particles produced by one formula unit of \(K_4[Fe(CN)_6]\) is:

\(4 \text{ (from \(K^+\))} + 1 \text{ (from \([Fe(CN)_6]^{4-}\))} = 5\)

So, the van't Hoff factor for \(K_4[Fe(CN)_6]\) is \(i = 5\).

Calculate the van't Hoff factors for the given substances

Now, let's determine the van't Hoff factors for each of the substances in the options:

1. \(AlCl_3\):

\(AlCl_3\) dissociates as:

\(AlCl_3 \longrightarrow Al^{3+} + 3Cl^-\)

Total particles: \(1 + 3 = 4\)

Van't Hoff factor, \(i = 4\)

2. \(AlN\):

\(AlN\) is a covalent compound that does not dissociate into ions in water. Therefore, it remains as one particle, and \(i = 1\).

3. \(AlF_3\):

\(AlF_3\) dissociates as:

\(AlF_3 \longrightarrow Al^{3+} + 3F^-\)

Total particles: \(1 + 3 = 4\)

Van't Hoff factor, \(i = 4\)

4. \(Al_2(SO_4)_3\):

\(Al_2(SO_4)_3\) dissociates as:

\(Al_2(SO_4)_3 \longrightarrow 2Al^{3+} + 3SO_4^{2-}\)

Total particles: \(2 + 3 = 5\)

Van't Hoff factor, \(i = 5\)

Compare the van't Hoff factors

\(K_4[Fe(CN)_6]\) has a van't Hoff factor of \(5\). Among the options, only \(Al_2(SO_4)_3\) has the same van't Hoff factor of \(5\).

Conclusion

Since both \(K_4[Fe(CN)_6]\) and \(Al_2(SO_4)_3\) dissociate to produce a total of 5 particles, they have the same van't Hoff factor. Therefore, the correct answer is option 4: \(Al_2(SO_4)_3\).