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: This question belongs to the topic of the colligative property of solutions. Since it is mentioned that the molality of all the given solutions is the same, i.e. 1 molal maybe, it tells us that we need to consider the Van’t Hoff factors for such solutions, which depends on the degree of association/dissociation of solute in the solution.

Complete step-by-step answer:
Colligative properties are the properties of a solution which solely depends on the number of particles of the solute and not on the type or nature of the solute. Freezing point of a solution is one such colligative property. Whenever we add a solute to the solution, it lowers the original freezing point of the solution and this depression in freezing point can be calculated with the help pf below formula:
 \[\Delta {T_f} = i \times {K_f} \times m\]
Where \[\Delta {T_f}\] is the freezing point depression, i is the Van’t Hoff factor, \[{K_f}\] is the freezing constant and m is the molality. Van’t Hoff factor, it depends on the degree of association/dissociation of solute in the solution and the number of ions produced in the solution. so, we can write this as
 \[\alpha = \dfrac{{i - 1}}{{n - 1}}\]
According to this, we get the information that for greater value of n, greater will be the Van’t Hoff factor, where n is number of ions produced on dissolution in a solvent. Let us check it for all the four compounds given to us.
For \[{C_6}{H_5}N{H_3}Cl\] , \[n = 3\]
For \[Ca{(N{O_3})_2}\] , \[n = 3\]
For \[La{(N{O_3})_3}\] , \[n = 4\]
For \[{C_6}{H_{12}}{O_6}\] , \[n = 1\]
The lowest Van’t Hoff factor will result in the lowest depression in freezing point and thus, the solution will have the highest freezing point. Since glucose has the lowest i, it will have the highest freezing point.
Hence the correct option is (D).

Note: Van’t Hoff factor comes into consideration only when the solution is an electrolyte, i.e. the solute which undergoes dissociation or association when present in an aqueous solution. Glucose is a monosaccharide and thus it cannot be dissociated further. So, it is a non-electrolyte and its Van’t Hoff factor is 1.