Out of $1 M B a C l_{2}$ and $1 M A l C l_{3}$ , which one has higher freezing point and why ?

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Solution

Depression in freezing point, $Δ = i (K_{f} m)$ ; where $i$ = Van't-Hoff factor
Here both $B a C l_{2}$ and $A l C l_{3}$ have same molarity, thus the freezing point will now depend upon the Van't Hoff's factor.
$B a C l_{2} ⇌ B a^{2 +} + 2 C l^{-}$ $i = 3$
$A l C l_{3} ⇌ A l^{3 +} + 3 C l^{-}$ $i = 4$
Therefore, $A l C l_{3}$ having more Van't Hoff factor, has higher freezing point than $1 M$ $B a C l_{2}$ .

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