The freezing point of the equimolal aqueous solution will be highest for:

\(C_6H_{12}O_6\) (glucose)

The correct answer is option 4. \(C_6H_{12}O_6\) (glucose).

Colligative properties are the properties of a solution that solely depend on the number of particles of the solute and not on the type or nature of the solute. The freezing point of a solution is one such colligative property. When a solute is added to a solution, it lowers the original freezing point of the solution. This depression in the freezing point can be calculated using the following formula:

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

where \(\Delta T_f\) is the freezing point depression, \(i\) is the Van't Hoff factor, \(K_f\) is the freezing constant, and \(m\) is the molality.

The Van't Hoff factor, \(i\), depends on the degree of association/dissociation of the solute in the solution and the number of ions produced in the solution. It can be calculated as:

\(\alpha = i - 1 + n - 1\)

According to this equation, for a greater value of \(n\), the Van't Hoff factor will be greater. Here, \(n\) represents the number of ions produced upon dissolution in a solvent. Let us calculate the Van't Hoff factor for the four compounds given to us:

For \(C_6H_5N^+H_3Cl^-\), \(n = 3\)

For \(Ca(NO_3)_2\), \(n = 3\)

For \(La(NO_3)_3\), \(n = 4\)

For \(C_6H_{12}O_6\), \(n = 1\)

The lowest Van't Hoff factor will result in the lowest depression in the freezing point, and thus, the solution will have the highest freezing point. Since glucose (\(C_6H_{12}O_6\)) has the lowest \(i\), it will have the highest freezing point.