Question

The freezing point of equimolal aqueous solution will be highest for:
A. ${{C}_{6}}{{H}_{5}}N{{H}_{3}}Cl$
B. $Ca{{(N{{O}_{3}})}_{2}}$
C. $La{{(N{{O}_{3}})}_{3}}$
D. ${{C}_{6}}{{H}_{12}}{{O}_{6}}$

Answer 1

Hint: Equimolal aqueous solution is a way of saying the molality of the two components which are in the solution, are equal. The value of freezing point is inversely proportional to the number of particles the solute will dissociate from one unit solute, or the van’t Hoff’s factor.

Formula used:
\[\vartriangle{{T}_{f}}=i\times{{K}_{f}}\times m\]
Here the $\vartriangle {{T}_{f}}$ represents depression in freezing point which is equal to ${{T}_{f}}-{{T}_{f}}'$, where both of them represents final and initial temperature of the solvent respectively.
$i$ represents the van’t Hoff’s factor, and the next symbol Kf represents the molal freezing point depression constant.
And $m$ represents the concentration in molality.

Complete answer:
Freezing point can be defined as the temperature at which the liquid state of a substance freezes to its solid state. There are different freezing points for different liquids. The depression in freezing point is a phenomena which can be observed when a solute is added to a solvent, as a result of which, the freezing point of that solvent decreases. When a solvent undergoes the process of freezing, the constituent particles of that solvent starts to slow down because of the decrease in temperature, as a result of which, the intermolecular forces between the molecules start to strengthen up. We know that when the constituent molecules feel strong attractions, they arrange themselves in such a pattern, that it becomes a solid. For instance, when water is cooled to its freezing point, the constituent molecules of water become slower and we know that the water has intermolecular hydrogen bonding between its molecules, so the constituent molecules will experience strong hydrogen bonds and it attains a solid form.
Now when the common salt is added to water, the ions of sodium and calcium get attracted to the water molecules, these attractive forces interfere with the solid formation of water. So know, to freeze the solution, we will have to freeze it to an even much lower temperature.
 \[\vartriangle {{T}_{f}}=i\times {{K}_{f}}\times m\]
Here the $\vartriangle {{T}_{f}}$ represents depression in freezing point which is equal to ${{T}_{f}}-{{T}_{f}}'$, where both of them represents final and initial temperature of the solvent respectively.
$i$ represents the van’t Hoff’s factor, which is the value which represents the number of particles formed from one unit of the solute, which is present in the solution.
And $m$ represents the concentration in molality, which is the number of moles present in per kilogram of solvent.
Now we can derive from this information that,
${{T}_{f}}\propto \dfrac{1}{i}$
This implies, that the freezing point of a solvent is inversely proportional to the van’t hoff’s factor. Now if we consider each of the options given in the question, ${{C}_{6}}{{H}_{5}}N{{H}_{3}}Cl$, $Ca{{(N{{O}_{3}})}_{2}}$, $La{{(N{{O}_{3}})}_{3}}$, ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ the molecule of glucose will have lowest value of $i$ as it is not an ionic molecule, and it will not dissociate in its constituent ions, when dissolved in water unlike the rest of them.
And since we know that the freezing point of a solvent is inversely proportional to the van’t Hoff’s factor. So, the glucose will have the highest freezing point.

So, the correct answer is “Option D”.

Note: The freezing point of a substance is inversely proportional to the Van’t Hoff’s factor, this means that with increase in freezing point, the value of van’t Hoff’s factor decreases, and vice versa.
The glucose molecule does not dissociate into its constituent ions, as it is not an ionic molecule, so it will have the minimum value of van’t Hoff’s factor among all of them. Hence, highest freezing point.