Chemistry - Liquid Solution Question with Solution | TestHub

As molar concentration is same so only deciding factor is vant Hoff factor
Depression in freezing point$\propto$vant Hoff's factor for$\text{A}{\text{l}}_2(\text{S}{\text{O}}_4{)}_3\to \mathrm{i}=5$
${\mathrm{K}}_2\text{S}{\text{O}}_4\to \mathrm{i}=3$
$\text{KCl}\to \mathrm{i}=2$
${\mathrm{C}}_6{\mathrm{H}}_{\text{12}}{\mathrm{O}}_6\to \mathrm{i}=1$