JEE Main 2021 — Solutions Question with Solution

To determine which solution has the lowest freezing point, you should consider the van 't Hoff factor $i$, which represents the number of particles the solute dissociates into when dissolved. The freezing point depression is given by:

$$\Delta T_f=i\cdot K_f\cdot m$$

where:

$\Delta T_f$ is the change in freezing point,

$K_f$ is the freezing point depression constant,

$m$ is the molality of the solution,

$i$ is the van 't Hoff factor, which depends on the dissociation of the solute.

For each solute, the van 't Hoff factor $i$ can be determined as follows:

Option A: ${\text{Al}}_2({\text{SO}}_4{)}_3$

Dissociates into 2 Al${}^{3+}$ ions and 3 SO${}_4^{2-}$ ions.

$i=2+3=5$

Option B: ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ (glucose)

Does not dissociate in solution (non-electrolyte).

$i=1$

Option C: $\text{KI}$

Dissociates into K${}^{+}$ and I${}^{-}$ ions.

$i=1+1=2$

Option D: ${\text{K}}_2{\text{SO}}_4$

Dissociates into 2 K${}^{+}$ ions and 1 SO${}_4^{2-}$ ion.

$i=2+1=3$

Comparing the van 't Hoff factors, the solution of ${\text{Al}}_2({\text{SO}}_4{)}_3$ will have the highest factor ($i=5$), leading to the greatest freezing point depression. Thus, the ${\text{Al}}_2({\text{SO}}_4{)}_3$ solution will have the lowest freezing point since the extent of freezing point depression is directly proportional to $i$.

Therefore, the solution of ${\text{Al}}_2({\text{SO}}_4{)}_3$ has the lowest freezing point.