Which one of the following solutions will exhibit smallest freezing point depression?

$0.1\, m\, C_6H_{12}O_6$

The correct answer is Option (2) → $0.1\, m\, C_6H_{12}O_6$.

The freezing point depression (\(\Delta T_f\)) depends on both the molality of the solution and the number of particles into which the solute dissociates in solution. The formula for freezing point depression is:

\(\Delta T_f = i \cdot K_f \cdot m\)

Where:

\(i\) is the van 't Hoff factor, which represents the number of particles the solute dissociates into.

\(K_f\) is the freezing point depression constant of the solvent (water in this case).

\(m\) is the molality of the solution.

Since all solutions have the same molality (0.1 m), we need to focus on the van 't Hoff factor (\(i\)) for each solute to determine which will exhibit the smallest freezing point depression.

1. \(K_2SO_4\) dissociates into 3 ions: \(2K^+\) and \(SO_4^{2-}\), so \(i = 3\).

2. \(C_6H_{12}O_6\) (glucose) does not dissociate in solution, so \(i = 1\).

3. \(Al_2(SO_4)_3\) dissociates into 5 ions: \(2Al^{3+}\) and \(3SO_4^{2-}\), so \(i = 5\).

4. \(KCl\) dissociates into 2 ions: \(K^+\) and \(Cl^-\), so \(i = 2\).

Conclusion:

The solution with the smallest freezing point depression will be the one with the smallest van 't Hoff factor (\(i\)), which is \(C_6H_{12}O_6\) (glucose) with \(i = 1\).

Thus, the solution that will exhibit the smallest freezing point depression is 0.1 m \(C_6H_{12}O_6\) (glucose),