Step 1: Recall freezing point depression formula.
\[ \Delta T_f = i \cdot K_f \cdot m \] Smaller \( i \times m \) means less depression and a higher freezing point.
Step 2: KCl solution.
KCl dissociates into \( K^+ + Cl^- \), so \( i = 2 \). Molality = 0.1 mol/kg. Effective \( i \times m = 0.2 \).
Step 3: K2SO4 solution.
\( K_2SO_4 \) dissociates into \( 2K^+ + SO_4^{2-} \), so \( i = 3 \). Molality = 0.1 mol/kg. Effective \( i \times m = 0.3 \).
Step 4: Urea solution.
Urea is a non-electrolyte, \( i = 1 \). Molality = 0.1 mol/kg. Effective \( i \times m = 0.1 \).
Step 5: Glucose solution.
Glucose molar mass = 180 g/mol. 30 g in 1 kg water gives molality = \( 30/180 = 0.167 \) mol/kg. Non-electrolyte, \( i = 1 \). Effective \( i \times m = 0.167 \).
Step 6: Identify the highest freezing point.
Comparing: urea gives the smallest \( i \times m = 0.1 \), so it has the least depression and the highest freezing point. \[ \boxed{0.1 \text{ mol urea in 1 kg water}} \]