Which one of the following 0.06 M aqueous solution has lowest freezing point?
(JEE MAIN 2021)

K I

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C_{6} H_{12} O_{6}

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K_{2} S O_{4}

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A l_{2} (S O_{4})_{3}

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Solution

The correct option is D $A l_{2} (S O_{4})_{3}$
$Δ T_{f} = i K_{f} m$ $m$ is molality $Δ T_{f} = T_{f}^{0} - T_{f}$ $i$ is van't Hoff factor $T_{f}^{0}$ is the freezing point of pure solvent $T_{f}$ is the freezing point of solution Here $Δ T_{f} \propto i$ (for equimolar solutions) For minimum $T_{f}$ , the value of i should be the highest $K I H_{2} O - - \to K^{+} (a q) + I^{-} (a q)$ $A l_{2} (S O_{4})_{3} H_{2} O - - \to 2 A l^{3 +} (a q) + 3 S O_{4}^{2 -} (a q)$ $For organic solvents like C_{6} H 12 O_{6}; i = 1$ $K_{2} (S O_{4}) H_{2} O - - \to 2 K^{+} (a q) + S O_{4}^{2 -} (a q)$
Summarising:

Solute $i$

$K I$ 2

$A l_{2} (S O_{4})_{3}$ 5

$C_{6} H_{12} O_{6}$ 1

$K_{2} S O_{4}$ 3

$A l_{2} (S O_{4})_{3}$ has the highest value of i and hence has lowest freezing point

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