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

C₆H₁₂O₆

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

To determine which equimolal aqueous solution will have the highest freezing point, we need to consider the effect of solute dissociation on freezing point depression. The key concept here is the van 't Hoff factor (\(i\)), which represents the number of particles into which a solute dissociates in solution. The greater the number of particles, the greater the freezing point depression.

The freezing point depression (\(\Delta T_f\)) is given by the formula:

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

where:

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

\(K_f\) is the cryoscopic constant.

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

Analysis of Each Solute

1. \(C_6H_5NH_3Cl\) (Phenylammonium chloride):

Dissociates into two ions: \( C_6H_5NH_3^+ \) and \( Cl^- \)

\(i = 2\)

2. \(Ca(NO_3)_2\) (Calcium nitrate):

Dissociates into three ions: \( Ca^{2+} \) and 2 \( NO_3^- \)

\(i = 3\)

3. \(La(NO_3)_3\) (Lanthanum nitrate):

Dissociates into four ions: \( La^{3+} \) and 3 \( NO_3^- \)

\(i = 4\)

4. \(C_6H_{12}O_6\) (Glucose):

Does not dissociate in water (non-electrolyte)

\(i = 1\)

Determining the Highest Freezing Point

The freezing point depression is inversely proportional to the van 't Hoff factor \(i\). Therefore, the solution with the smallest van 't Hoff factor will experience the least freezing point depression and thus have the highest freezing point.

Since glucose \((C_6H_{12}O_6)\) has the smallest van 't Hoff factor (\(i = 1\)), its solution will have the least freezing point depression and therefore the highest freezing point among the given options.