Decreasing order of freezing point of 0.1 M aqueous solution of following will be:

$\begin{array}{cc}\left(A\right)& K_4\left[\mathrm{Fe}\right(\mathrm{CN}\left)6\right]\\ \left(B\right)& K_2{\mathrm{SO}}_4\\ \left(C\right)& {\mathrm{NH}}_2\mathrm{CONH}2\\ \left(D\right)& \mathrm{AlCl}3\\ \left(E\right)& \mathrm{HCl}\end{array}$

Choose the correct answer from the options given below:

The correct option is (A) > (D) > (B) > (E) > (C).

Explanation:

The depression in freezing point (
$\Delta T_f$
) is a colligative property, meaning it depends on the number of solute particles in the solution. It is defined by the formula:

$\Delta T_f=i·K_f·m$

where:
• $i$ is the van 't Hoff factor (representing the number of ions/particles after dissociation)
• $K_f$ is the molal depression constant of the solvent
• $m$ is the molality of the solution

Since the concentration for all solutions is identical (0.1 M), the depression in freezing point is directly proportional to the van 't Hoff factor ($i$):

$\Delta T_f\propto i$

Let's calculate the value of $i$ for each compound assuming complete dissociation:

(A) $K_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]$:
This complex salt dissociates into 4 potassium ions and 1 ferrocyanide complex ion:
$K_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]\to 4K^{+}+[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$
So, $i=4+1=5$.

(D) ${\mathrm{AlCl}}_3$:
This ionic compound dissociates into 1 aluminum ion and 3 chloride ions:
${\mathrm{AlCl}}_3\to {\mathrm{Al}}^{3+}+3{\mathrm{Cl}}^{-}$
So, $i=1+3=4$.

(B) $K_2{\mathrm{SO}}_4$:
This salt dissociates into 2 potassium ions and 1 sulfate ion:
$K_2{\mathrm{SO}}_4\to 2K^{+}+{\mathrm{SO}}_4^{2-}$
So, $i=2+1=3$.

(E) $\mathrm{HCl}$:
This strong acid dissociates into 1 hydrogen ion and 1 chloride ion:
$\mathrm{HCl}\to H^{+}+{\mathrm{Cl}}^{-}$
So, $i=1+1=2$.

(C) ${\mathrm{NH}}_2{\mathrm{CONH}}_2$ (Urea):
Urea is a covalent organic compound that does not dissociate in water.
So, $i=1$.

Arranging these solutions in decreasing order of their van 't Hoff factor ($i$) and consequently the freezing point depression gives:
(A) > (D) > (B) > (E) > (C)