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Question
assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing points. highest freezing point nh₄i cobr₃(aq) na₂co₃(aq) lowest freezing point answer bank
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 $NH_4I
ightarrow NH_4^++I^-$, so $i = 2$. For $Na_2CO_3$, it dissociates as $Na_2CO_3
ightarrow 2Na^++CO_3^{2 - }$, so $i = 3$. For $CoBr_3$, it dissociates as $CoBr_3
ightarrow Co^{3+}+3Br^-$, 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.
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Highest freezing point: $NH_4I$
Next: $Na_2CO_3$
Lowest freezing point: $CoBr_3$