Question

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

• A. $[Co(H_2O)6]Cl_3 $
• B. $[Co(H_2O)_5Cl]Cl_2.H_2O$
• C. $[Co(H_2O)_4Cl_2]Cl._2H_2O $
• D. $[Co(H_2O)_3Cl_3]._3H_2O $

Answer 1

The Correct Option is D

To determine which compound has the highest freezing point among the given options, we need to apply the concept related to the colligative properties of solutions, specifically the freezing point depression. The freezing point of a solution is lowered in comparison to that of the pure solvent, and this lowering is proportional to the molality of the solution and the number of particles the solute breaks into when dissolved.

The formula for freezing point depression is:

\Delta T_f = i \cdot K_f \cdot m

Where:

• \Delta T_f is the freezing point depression.
• i is the van’t Hoff factor (number of particles the compound dissociates into).
• K_f is the cryoscopic constant of the solvent.
• m is the molality of the solution.

The compound with the lowest van't Hoff factor will show the highest freezing point, as it causes the least depression.

Let's examine each compound:

1. [Co(H_2O)_6]Cl_3: Dissociates into 4 ions: 1 complex ion and 3 chloride ions (i = 4).
2. [Co(H_2O)_5Cl]Cl_2.H_2O: Dissociates into 3 ions: 1 complex ion and 2 chloride ions (i = 3).
3. [Co(H_2O)_4Cl_2]Cl._2H_2O: Dissociates into 3 ions: 1 complex ion and 1 chloride ion + additional non-dissociating water (i = 3).
4. [Co(H_2O)_3Cl_3]._3H_2O: Does not dissociate into ions, as all chlorides are inside the coordination sphere (i = 1).

From this analysis, we find that [Co(H_2O)_3Cl_3]._3H_2O has the lowest van't Hoff factor of 1 because it does not dissociate into ions. Therefore, it will show the least freezing point depression and thus, the highest freezing point among the options.

Hence, the correct answer is [Co(H_2O)_3Cl_3]._3H_2O.