The freezing point of equimolal aqueous solution will be highest for:

C_{6} H_{5} {N H}_{3}^{+} {C l}^{-}

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C a {({N O}_{3})}_{2}

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L a {({N O}_{3})}_{2}

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

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Solution

The correct option is B $C_{6} H_{12} O_{6}$
Freezing point $\propto \frac{1}{Van't Hoff factor}$

For $C_{6} H_{12} O_{6} i$ is minimum

So, its freezing point will be highest.
Depression in freezing point, $Δ T_{f} = i K_{f} m$

The value of van't Hoff factor (i) is minimum for the glucose, which is a non-electrolyte. Hence, aqueous solution of glucose has highest freezing point.

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Similar questions

$18.0 g$ of glucose $(C_{6} H_{12} O_{6})$ is added to $178.2 g$ of water. The vapour pressure of water for this aqueous solution at $100^{o} C$ is :

{Δ H}_{f}^{o}

C_{6} H_{12} O_{6} (s)

Δ H_{c o m b .}

C_{6} H_{12} O_{6} (s) = - 2816 k J {m o l}^{- 1}

{Δ H}_{f}^{o}

{C O}_{2} (g) = - 393.5 J {m o l}^{- 1}

{Δ H}_{f (H_{2} O)}^{o} = - 285.9 k J {m o l}^{- 1}

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

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