Answer:

Correct Answer: C

Solution:

$ \Delta T _{f} $ (freezing point depression) is a colligative property and depends upon the van’t Hoff factor (i)

i.e., number of ions given by the electrolyte in aqueous solution.

$ \Delta T _{f}=i\times K _{f}\times m $ where, $ K _{f} $ = molal freezing point depression constant. m = molality of the solution

$ \therefore $ $ K _{f} $ and m are constant, $ \Delta T _{f}\propto i $

[a] $ KCl(aq){K^{+}}(aq)+C{l^{-}}(aq), $ (Total ions = 2 thus, i = 2)

[b] $ C _6H _{12}O _6no,ions[i=0] $

[c] $ Al _2{{(SO _4)}_3}(aq)2A{l^{3+}}+3SO_4^{2-} $ [Total ions =5, thus, i = 5]

[d] $ K _2SO _4(aq)2{K^{+}}+SO_4^{2-} $ [Total ions = 3, t Thus, i = 3]

Hence, $ Al _2{{(SO _4)}_3} $ will exhibit largest freezing point depression due to the highest value of i.