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

Explanation:

Step1: Recall freezing - point depression formula

The freezing - point depression $\Delta T_f = iK_fm$, where $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant (constant for a given solvent), and $m$ is the molality. Since concentrations are equal, the solution with the lowest $i$ will have the highest freezing point.

Step2: Determine van't Hoff factors

For $NaNO_3$, it dissociates as $NaNO_3\rightarrow Na^++NO_3^-$, so $i = 2$. For $K_2SO_4$, it dissociates as $K_2SO_4\rightarrow 2K^++SO_4^{2 - }$, so $i = 3$. For $K_3PO_4$, it dissociates as $K_3PO_4\rightarrow 3K^++PO_4^{3 - }$, so $i = 4$. For $Al_2(SO_4)_3$, it dissociates as $Al_2(SO_4)_3\rightarrow 2Al^{3+}+3SO_4^{2 - }$, so $i = 5$.

Step3: Arrange by freezing point

The higher the $i$ value, the lower the freezing point. So the order from highest to lowest freezing point is $NaNO_3(aq)>K_2SO_4(aq)>K_3PO_4(aq)>Al_2(SO_4)_3(aq)$.

Answer:

Highest freezing point: $NaNO_3(aq)$ $K_2SO_4(aq)$ $K_3PO_4(aq)$ Lowest freezing point: $Al_2(SO_4)_3(aq)$