Question

Which will have the lowest freezing point?
A) A 0.01 molar solution of $HN{O_3}$
B) A 0.01 molar solution of $H{C_2}{H_3}{O_2}$
C) A 0.01 molar solution of $Cu{(N{O_3})_2}$
D) A 0.01 molar solution of $NaN{O_3}$

Answer 1

Hint: Recall the van’t Hoff factor ($i$ ), which is the number of discrete ions formed in a solution from one formula unit of solute. Van’t hoff factor also accounts for the effect of a solute on colligative properties (such as, freezing point depression, boiling point elevation etc.). The compound for which $i$ is maximum, will have the maximum depression in freezing point.

Complete step by step answer:
We know that when ionic compounds dissolve in water, they dissociate into cations and anions. In 1880, van’t Hoff introduced a factor $i$, known as the van’t Hoff factor, to account for the extent of association or dissociation. It is also the measure of effect of a solute on colligative properties. Depression in freezing point is one of the colligative properties. After inclusion of van’t Hoff factor $i$ in equation of freezing-point depression colligative property:
$\Delta {T_f} = i{\text{ }}{k_f}m$
Where, $\Delta {T_f}$ is depression in freezing point, ${k_f}$ is the molal depression constant, $m$ is the molality of the solution.

Or, we can say, depression in freezing point is directly proportional to the van’t Hoff factor, i.e. $\Delta {T_f}{\text{ }}\alpha {\text{ }}i$
The ionic compounds given in options have the same concentration of 0.01 molar. Now, calculating $i$ for each ionic compound:
- Dissociation of $HN{O_3}$:
$HN{O_3} \to {H^ + } + NO_3^ - $
$HN{O_3}$ dissociates in water to give one hydrogen and one nitrate ion. Therefore, $i$ for $HN{O_3}$ is 2.

 Dissociation of $H{C_2}{H_3}{O_2}$ or $C{H_3}COOH$:
$C{H_3}COOH \to {H^ + } + C{H_3}CO{O^ - }$
Similarly, $i$ for $C{H_3}COOH$ is equal to 2.

- Dissociation of $Cu{(N{O_3})_2}$:
$Cu{(N{O_3})_2} \to C{u^{2 + }} + 2NO_3^ - $
Therefore, $i$ for $Cu{(N{O_3})_2}$ = 3, since total three ions are formed on dissociation.

- Dissociation of $NaN{O_3}$:
$NaN{O_3} \to N{a^ + } + NO_3^ - $
Hence, $i$ for $NaN{O_3}$ is 2.
Therefore, $Cu{(N{O_3})_2}$ has the highest value of van’t Hoff factor, i.e. 3, among all ionic compounds. Consequently, $Cu{(N{O_3})_2}$ will have the maximum depression in freezing point and hence will have the lowest freezing point.
So, the correct answer is “Option C”.

Note: For solutes showing association, $i$ > 1
For solutes showing dissociation, $i$ < 1
For solutes showing no association or dissociation (non-electrolytes), $i$= 1.