Question

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

• A. \( [\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_3 \)
• B. \( [\text{Co}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O} \)
• C. \( [\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl}\cdot2\text{H}_2\text{O} \)
• D. \( [\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3]\cdot3\text{H}_2\text{O} \)

Hint:

The freezing point of a solution increases as the number of dissociated particles decreases. Lower dissociation means fewer particles, leading to a higher freezing point.

Answer 1

The Correct Option is D

Step 1: Freezing Point Depression Formula The freezing point depression is directly related to the number of dissociated particles in the solution. The formula is:

$$ \Delta T_f = K_f \cdot m \cdot i $$

where:

• $ \Delta T_f $ is the freezing point depression,
• $ K_f $ is the cryoscopic constant (which depends on the solvent),
• $ m $ is the molality of the solution,
• $ i $ is the van 't Hoff factor, representing the number of particles resulting from dissociation.

Since the freezing point depression is inversely proportional to the number of dissociated particles, the solution with the lowest $ i $ (fewer particles) will have the highest freezing point.

Step 2: Identify the $ i $ for Each Compound Let's now identify the van 't Hoff factor $ i $ for each compound:

1. $ [\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_3 $ dissociates into 4 particles: $ [\text{Co}(\text{H}_2\text{O})_6]^{3+} $ and 3 $ \text{Cl}^- $. Hence, $ i = 4 $.
2. $ [\text{Co}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O} $ dissociates into 3 particles: $ [\text{Co}(\text{H}_2\text{O})_5\text{Cl}]^{2+} $ and 2 $ \text{Cl}^- $. Hence, $ i = 3 $.
3. $ [\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O} $ dissociates into 2 particles: $ [\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]^{+} $ and $ \text{Cl}^- $. Hence, $ i = 2 $.
4. $ [\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O} $ does not dissociate, so it remains as a single particle: $ i = 1 $.

Step 3: Conclusion Since the freezing point depression is inversely proportional to $ i $, the compound with the lowest $ i $ will have the highest freezing point. In this case, $ [\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O} $, which has $ i = 1 $, will have the highest freezing point.

Final Answer:

$$ \boxed{[\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3]\cdot3\text{H}_2\text{O}} $$