For 0.1 M solutions of urea and Al_2(SO_4)_3 | vedclass

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Question

(D) Urea is a non-electrolyte,so its van't Hoff factor $(i)$ is $1$.
$Al_2(SO_4)_3$ is an electrolyte that dissociates as $Al_2(SO_4)_3 \rightarr

Vedclass · Accepted Answer

$a, b, c$ and $d$

Vedclass · Answer

(D) Urea is a non-electrolyte,so its van't Hoff factor $(i)$ is $1$.
$Al_2(SO_4)_3$ is an electrolyte that dissociates as $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so its van't Hoff factor $(i)$ is $5$.
Colligative properties like elevation in boiling point $(\Delta T_b = i K_b m)$ and depression in freezing point $(\Delta T_f = i K_f m)$ are directly proportional to $i$.
Since $i_{Al_2(SO_4)_3} > i_{urea}$,$Al_2(SO_4)_3$ will have a higher elevation in boiling point and a higher depression in freezing point.
Lower depression in freezing point means a higher freezing point for urea compared to $Al_2(SO_4)_3$.
Lower solute particles mean higher vapor pressure for urea compared to $Al_2(SO_4)_3$.
Thus,statements $a, b, c,$ and $d$ are correct.

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