QJEEM18Ch2fc0: JEE Main 2018 Chemistry - Class 12 JEEM18Ch2fc0 Solution: For 1 molal aqueous solution of the following compounds, which…

We know,

$\Delta {T_f}\, = \,im{K_f}$

Where

$\Delta {T_f}\, = \,$ Depression in freezing point

i = no of ions or molecule

m = molality of solute

K_f = freezing point depression constant.

In this question m = 1 and K_f is constant.

So, $\Delta {T_f}\,\, \propto \,\,i$

Which solutions have more i, those solutions $\Delta {T_f}\,$ will be more.

Now,

Freezing point of a aqueous solution ${f_p} = \, - \,\Delta {T_f}$

So, those solutions have more $\Delta {T_f}$ will have less freezing point

and $\Delta {T_f}$ will be more when value of i is more.

Now, for,

[ Co (H₂O)₆] Cl₃ $\rightleftharpoons$ [ Co(H₂O)₆]³⁺ + 3Cl ^$-$

Value of i = 4 (no. of ions), 3 ions of Cl ^$-$

and one ion of [ Co (H₂O)₆]³⁺

[ Co (H₂O)₅ Cl ] Cl₂ . H₂O $\rightleftharpoons$ [ Co (H₂O)₅ Cl ]²⁺ + 2Cl ^$-$

Value of i = 3

[ Co (H₂O)₄ Cl₂ ] Cl . 2H₂O $\rightleftharpoons$ [ Co (H₂O)₄ Cl₂ ] ⁺ + Cl ^$-$

Value of i = 2

and [ Co(H₂O)₃ Cl ] . 3H₂O do not show ionization.

So, i = 1.

As, [ Co (H₂O)₃ Cl ]. 3H₂O have lowest value of i,

So, it's freezing point will be maximum.