The freezing point of equimolal aqueous solution will be highest for which of the following?

C₆H₁₂O₆

The correct answer is option 4. C₆H₁₂O₆

The freezing point of an equimolal aqueous solution will be highest for the solution with the lowest number of particles.

Colligative properties, like freezing point depression, depend on the number of solute particles that are not attracted to the solvent molecules (free in the solution). More solute particles disrupt the solvent's crystal lattice formation, leading to a lower freezing point.

Let's analyze the options:

\(C_6H_5NH_4Cl\) (aniline hydrochloride): This dissociates into two ions (\(C_6H_5NH_4^+\) and \(Cl^-\)) in water.

\(Ca(NO_3)_2\) (calcium nitrate): This dissociates into three ions (\(Ca^{2+}\) and \(2NO_3^-\)) in water.

\(La(NO_3)_3\) (lanthanum nitrate): This dissociates into four ions (\(La^{3+}\) and \(3NO_3^-\)) in water.

\(C_6H_{12}O_6\) (glucose): Glucose is a covalent molecule and doesn't dissociate significantly in water. It will have the fewest total particles (one molecule) compared to the ionic compounds that break into multiple ions.

Therefore, the freezing point of the equimolal aqueous solution of \(C_6H_{12}O_6\) (glucose) will be the highest because it has the fewest particles in solution.