A) Sucrose \[>KCl>BaC{{l}_{2}}>AlC{{l}_{3}}\]
B) \[AlC{{l}_{3}}>BaC{{l}_{2}}>KCl>\] Sucrose
C) \[BaC{{l}_{2}}>KCl>AlC{{l}_{3}}>\] Sucrose
D) \[KCl>BaC{{l}_{2}}>AlC{{l}_{3}}>\]Sucrose
Correct Answer: A
Solution :
Depression in freezing point is a colligative property, it means that as the number of particles (ions) of solute increases, freezing point of the solution decreases. Sucrose \[\xrightarrow{{}}\]not ionised \[KCl\rightleftharpoons \underbrace{{{K}^{+}}+C{{l}^{-}}}_{2\,\,ions}\] \[BaC{{l}_{2}}\rightleftharpoons \underbrace{B{{a}^{2+}}+2C{{l}^{-}}}_{3\,\,ions}\] \[AlC{{l}_{2}}\rightleftharpoons \underbrace{A{{l}^{3+}}+3C{{l}^{-}}}_{4\,\,ions}\] Hence, the correct order of freezing point is Sucrose \[>KCl>BaC{{l}_{2}}>AlC{{l}_{3}}\]
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