Which of the following aqueous solutions has the lowest freezing point? — Solutions Chemistry Question

STEPS:

1. Identify the Underlying Chemistry Concept: This question tests the understanding of colligative properties, specifically freezing point depression. Colligative properties depend only on the number of solute particles in a solution, not their identity.
2. Apply the Freezing Point Depression Formula: The change in freezing point ($\Delta T_f$) is calculated using the equation:
$$\Delta T_f = i \cdot K_f \cdot m$$
where $i$ is the van't Hoff factor (the number of particles the solute dissociates into), $K_f$ is the molal freezing-point depression constant of the solvent, and $m$ is the molality of the solution.
3. Evaluate Constant Variables: In this problem, all solutions have the same molality (0.2 m) and the same solvent (water). Therefore, $K_f$ and $m$ are constant for all choices.
4. Relate the Van't Hoff Factor to Freezing Point: Since $\Delta T_f$ is proportional to $i$, the solution with the largest $i$ value (most particles) will have the greatest freezing point depression. A greater depression means the freezing point is lowered the most, resulting in the lowest freezing point.
5. Determine the Value of $i$ for Each Solute:
* NaCl: Dissociates into 2 ions ($Na^+$ and $Cl^-$). $i = 2$
* $CaCl_2$: Dissociates into 3 ions ($Ca^{2+}$ and $2 Cl^-$). $i = 3$
* $H_2SO_4$: A strong acid that dissociates into 3 ions ($2 H^+$ and $SO_4^{2-}$). $i = 3$
* $NH_3$: A weak base that only partially ionizes in water, remaining mostly as molecules. $i \approx 1$
* $Al(NO_3)_3$: Dissociates into 4 ions ($1 Al^{3+}$ and $3 NO_3^-$). $i = 4$
6. Conclusion: Because $Al(NO_3)_3$ produces the greatest number of particles per formula unit ($i=4$), its $0.2\text{ m}$ solution will have the largest total concentration of particles ($0.8\text{ m}$) and thus the lowest freezing point. This matches option E.

WHY_OTHERS_WRONG:

• A) 0.2 m NaCl: This solution only produces $0.4\text{ moles}$ of particles per kilogram of solvent ($i=2$). It will not lower the freezing point as much as the other ionic salts.
• B) 0.2 m $CaCl_2$ and C) 0.2 m $H_2SO_4$: Both of these solutes produce 3 particles per formula unit ($i=3$), resulting in a particle concentration of $0.6\text{ m}$. While they lower the freezing point more than NaCl, they do not lower it as much as $Al(NO_3)_3$.
• D) 0.2 m $NH_3$: Ammonia is a weak electrolyte (weak base) and does not dissociate significantly. It produces the fewest particles ($\approx 0.2\text{ m}$) and will therefore have the highest freezing point among the choices.