Animated Solution for Chemistry - Solutions: Which one of the following 0.06 M aqueous solutions has lowest freezing point? [2021, 22 July Shift-II]

Imagine you are standing outside on a freezing winter day, watching salt being thrown onto icy roads. Have you ever wondered why that works? The answer lies in a fascinating realm of physical chemistry known as colligative properties. These are properties of solutions that depend strictly on the number of solute particles present, completely ignoring their chemical identity.

When you dissolve a solute in a pure solvent like water, you disrupt its pristine, orderly structure. The entropy (or randomness) of the liquid state increases. Because the liquid is now more chaotic, it requires an even lower temperature to force those molecules into the highly ordered, rigid structure of a solid crystal lattice. This phenomenon is called depression in freezing point.

The mathematical relationship governing this drop in temperature is elegantly simple:

Here, $\Delta T_f$ is the depression in freezing point (how much the temperature drops), $K_f$ is the molal depression constant (a fixed value for the solvent), and $m$ is the molality of the solution.

But the real star of the show is $i$, the van't Hoff factor. Introduced by the brilliant Dutch chemist Jacobus Henricus van 't Hoff, this factor accounts for the fact that some substances break apart into multiple pieces when dissolved. If a molecule splits into three ions, it contributes three times as many particles to the solution, tripling the freezing point depression!

In our specific problem, we are given four different aqueous solutions. The beautiful part? They all have the exact same concentration: $0.06\text{ M}$. Since the solvent is water for all of them, $K_f$ is also constant. Therefore, the equation simplifies dramatically. The depression in freezing point becomes directly proportional to the van't Hoff factor:

To find the solution with the lowest freezing point, we need to find the one that causes the greatest depression. In other words, we are hunting for the compound that produces the maximum number of particles. Let's evaluate our contenders one by one.

1. Glucose (${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$) Glucose is a covalent compound. When you stir it into water, it dissolves, but it does not dissociate into ions. It remains as intact, neutral molecules. Therefore, one molecule of glucose yields exactly one particle in solution.

2. Potassium Iodide ($\text{KI}$) Potassium iodide is a strong electrolyte. The moment it hits the water, the ionic bonds shatter, and it completely dissociates into two distinct ions: a potassium cation and an iodide anion.

Since one formula unit gives two particles, its van't Hoff factor is:

3. Potassium Sulfate (${\text{K}}_2{\text{SO}}_4$) Things are getting more crowded. Potassium sulfate is also a strong electrolyte. Upon dissociation, it releases two potassium ions and one sulfate ion.

That gives us a total of three independent particles swimming around.

4. Aluminum Sulfate (${\text{Al}}_2({\text{SO}}_4{)}_3$) Finally, we reach the heavyweight champion. Aluminum sulfate breaks down to release two highly charged aluminum ions and three sulfate ions.

That is a massive total of five particles generated from just a single formula unit!

Let's bring it all together. We established that the depression in freezing point, $\Delta T_f$, is directly proportional to the van't Hoff factor, $i$.

Because aluminum sulfate (${\text{Al}}_2({\text{SO}}_4{)}_3$) has the highest van't Hoff factor ($i=5$), it will cause the most extreme disruption to the water's structure. It will produce the maximum depression in freezing point.

Now, don't make a silly mistake here! A maximum depression means the temperature drops the furthest. Therefore, the actual freezing point ($T_f$) of the aluminum sulfate solution will be the lowest among all the options.

This is a classic, high-yield concept for competitive exams. But what if the examiner decides to twist the knife? What if the concentrations were not equal?

If you are given solutions with varying molalities, you can no longer just look at the van't Hoff factor. You must calculate the effective concentration of particles for each solution by multiplying $i$ by $m$ ($i\times m$). The solution with the highest $i\times m$ product will have the lowest freezing point. Always remember the golden rule of colligative properties: It is all about the headcount!