assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing…

Explanation:

Step1: Recall colligative - property formula

The freezing - point depression formula is $\Delta T_f = iK_fm$, where $\Delta T_f$ is the change in freezing point, $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant (constant for a given solvent, here water), and $m$ is the molality. Since concentrations are equal ($m$ is the same) and $K_f$ is constant for water, the freezing - point depression depends only on the van't Hoff factor $i$.

Step2: Determine van't Hoff factors

For $NH_4I$, it dissociates as $NH_4I\rightarrow NH_4^++I^-$, so $i = 2$. For $Na_2CO_3$, it dissociates as $Na_2CO_3\rightarrow 2Na^++CO_3^{2 - }$, so $i = 3$. For $CoBr_3$, it dissociates as $CoBr_3\rightarrow Co^{3+}+3Br^-$, so $i = 4$.

Step3: Relate van't Hoff factor to freezing point

The higher the $i$ value, the greater the freezing - point depression $\Delta T_f$. A greater $\Delta T_f$ means a lower freezing point. So, the solution with the lowest $i$ value will have the highest freezing point.

Answer:

Highest freezing point: $NH_4I$ Next: $Na_2CO_3$ Lowest freezing point: $CoBr_3$