Question

The freezing-point depression of a 0.091 m solution of CsCl is

0.320C. The freezing-point depression of a 0.091 m solution of CaCl2 is

0.440 C. In which solution does ion association appear to be greater? Explain.

Short answer

The solution with greater ion association is${\mathrm{CaCI}}_2$.

Step-by-step solution

Step 1:

Freezing point depression = $\Delta T=i\times m\times K$

where K is the molal freezing constant or molal boiling constant of the solvent

$i=\frac{\Delta T}{m\times K}$

K is the same for both the solutions since the solvent is the same.

Step 2:

Expected Van’t Hoff factor for CsCl is 2 since it dissociates into 1 Cs cation

and 1 Cl anion.

$K_{CsCl}=\frac{\Delta T}{i\times m}$=$\frac{0.320}{2\times 0.09}=1.{77}^{0}Ckgmo{l}^{-1}$

Expected Van’t Hoff factor for ${\mathrm{CaCI}}_2$is 3, since it dissociates into 1 Ca

cation and 2 Cl anion.

$K_{CaC{l}_2}=\frac{\Delta T}{i\times m}$=$\frac{0.320}{3\times 0.09}=1.{18}^{0}Ckgmo{l}^{-1}$

observed Van’t Hoff factor is given by

$i=\frac{\Delta T}{m\times K}$ [ solvent is the same in both cases].

For CsCl: $i=\frac{0.320}{0.091\times 1.77}=1.986$

So the deviation in i for CsCl is = 2-1.986 = 0.014.

This decreased value of i is due to ion association

For ${\mathrm{CaCI}}_2$: $i=\frac{0.320}{0.091\times 1.18}=2.980$

So, the deviation in i for ${\mathrm{CaCI}}_2$ is = 3-2.980 = 0.020.

Since the deviation in ${\mathrm{CaCI}}_2$is greater hence the ion association in ${\mathrm{CaCI}}_2$is

greater.