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Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points; CoCl3, K2CO3, NH4I

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With the complete dissociation, CoCl3 will dissociate into 4 different ions, K2CO3 will dissociate into 3 different ions, and lastly NH4I will dissociate into 5 different ions. Since freezing point depression is directly proportional to number of ions, the ranking would be:
                                      NH4I, CoCl3, K2CO3

Based on greatest depression to lowest depression. 

Answer: The order of increasing freezing points is  [tex]NH_4I<K_2CO_3<CoCl_3[/tex]

Explanation:

The expression of depression in freezing point is given as:

[tex]\Delta T_f=i\times k_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] = Depression in freezing point

i = Van't Hoff factor

[tex]k_f[/tex] = freezing point constant

m = molality

Let the molarity of the given solutions is '1 m'

For the given options:

  • Option 1:  1 m [tex]CoCl_3[/tex]

Value of i = 4

So, molal concentration will be = [tex](1\times 4)=4m[/tex]

  • Option 2:  1 m [tex]K_2CO_3[/tex]

Value of i = 3

So, molal concentration will be = [tex](1\times 3)=3m[/tex]

  • Option 3:  1 m [tex]NH_4I[/tex]

Value of i = 2

So, molal concentration will be = [tex](1\times 2)=2m[/tex]

As, the molal concentration of [tex]CoCl_3[/tex] is the highest, so its freezing point will be the highest.

Hence, the order of increasing freezing points follows:

[tex]NH_4I<K_2CO_3<CoCl_3[/tex]

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