Question

Which of the following solutions has the lowest freezing point temperature?
A.$0.1{\text{M }}MgC{l_2}$
B.$0.1{\text{M}}\,HCl{O_4}$
C.$0.1{\text{M}}\,N{H_4}OH$
D.$0.1{\text{M}}\,KOH$
E.$0.1{\text{M}}\,LiN{O_3}$

Answer 1

To answer this question, you should recall the concept of lowering of the freezing point of a solvent on the addition of a non-volatile solute as well as basic points about colligative properties. The compound which dissociates the most will have maximum depression in freezing point.

Complete step by step solution:
We know that the term Freezing point depression refers to the lowering of the freezing point of solvents upon the addition of solutes. Colligative properties depend on the presence of dissolved particles and their number but not on the identity of the dissolved particles.
The depression in the freezing point of a solution can be described by the following formula:
$\Delta {T_F}\; = \;{K_F}\; \times \;b\; \times \,i$
where: $\Delta {T_F}$ is the freezing-point depression, ${K_F}\;$ is cryoscopic constant, $b$ is the molality and $i$ is the van 't Hoff factor.
Upon the addition of a solute which is non-volatile, the vapour pressure of the solution is found to be lower than the vapour pressure of the pure solvent. This causes the solid and the solution to reach equilibrium at lower temperatures. As depression in freezing point is a colligative property, therefore it is dependent on the number constituent particles. Since all of them have the same molar concentration, the compound breaking up into the most individual particles will have maximum depression freezing point.
A. $MgC{l_2} \to M{g^{2 + }} + 2C{l^{1 - }}$ (3 particles)
B. $HCl{O_4} \to {H^{1 + }} + Cl{O_4}^{1 - }$ (2 particles)
C. $N{H_4}OH \to N{H_4}^{1 + } + O{H^{1 - }}$ (2 particles)
D. $KOH \to {K^{1 + }} + O{H^{1 - \;}}$ (2 particles)
E. $LiN{O_3} \to L{i^{1 + }} + N{O_3}^{1 - }\;$ (2 particles)
$MgC{l_2}$ produces the most number of particles and hence, it has the highest Van’t Hoff factor. Thus, it has the maximum depression in freezing point.

Therefore, we can conclude that the correct answer to this question is option A.

Note:
The Van’t Hoff factor which is an important aspect of colligative properties is defined as the ratio of the concentration of particles formed when a substance is dissolved to the concentration of the substance by mass. In the case of non-electrolytic substances which do not dissociate in water, the value of $i$ is generally 1. But in the case of ionic substances, due to dissociation, the value of $i$ is equal to the total number of ions present in one formula unit of the substance.