Which one of the following will have the highest value of van’t Hoff factor, \(i\)?

\(K_2SO_4\)

The correct answer is option 1. \(K_2SO_4\).

The van't Hoff factor (\(i\)) represents the number of ions produced in a solution per formula unit of a solute that dissociates or ionizes. It is determined by the extent of dissociation of a solute in a solution.

Let us analyze the dissociation of each compound:

1. \(K_2SO_4\) (potassium sulfate):
Dissociation: \(K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}\)
Van't Hoff factor (\(i\)): 3

2. \(MgSO_4\) (magnesium sulfate):
Dissociation: \(MgSO_4 \rightarrow Mg^{2+} + SO_4^{2-}\)
Van't Hoff factor (\(i\)): 2

3. \(KCl\) (potassium chloride):
Dissociation: \(KCl \rightarrow K^+ + Cl^-\)
Van't Hoff factor (\(i\)): 2

4. \(NaCl\) (sodium chloride):
Dissociation: \(NaCl \rightarrow Na^+ + Cl^-\)
Van't Hoff factor (\(i\)): 2

Therefore, among the given compounds, \(K_2SO_4\) has the highest value of the van't Hoff factor (\(i\)), which is 3.