Question

Which of the following has the highest freezing point?
A. 1m NaCl solution
B. 1m KCl solution
C. 1m $AlC{l_3}$ solution
D. 1m \[{C_6}{H_{12}}{O_6}\] solution

Answer 1

Hint: This is a question from the colligative property of solutions. Since the molality of all the given solutions is the same, that is, 1 molal, it gives us an idea that we will have to consider the Van’t Hoff factors for these solutions.

Complete answer:
Colligative properties are the properties of a solution which solely depend on the number of particles of the solute and not on the nature or type of solute.
Freezing point of a solution is also a colligative property. Whenever a solute is added to a solution, it lowers its original freezing point. The depression in freezing point can be calculated with the formula:
$\Delta {T_s} = i \times {K_f} \times m$
Where, ‘i’ is the Van’t Hoff factor which depends on the degree of association/dissociation of solute in the solution and number of ions produced in the solution:
\[\alpha = \dfrac{{i - 1}}{{n - 1}}\]
Greater the value of n, greater is the van’t hoff factor.
- For NaCl, n=2
- For KCl, n=2
- For $AlC{l_3}$ , n=4
- For \[{C_6}{H_{12}}{O_6}\], n=1

The lowest van’t hoff factor will result in the least depression in freezing point and hence that solution will have the highest freezing point.

Hence, the correct option is D, glucose.

Note: Van’t Hoff factor only comes into play when the solute is an electrolyte, that is, the solute which dissociates or associates when in an aqueous solution. Glucose is a monosaccharide which cannot be dissociated further. Hence is a non electrolyte and its van’t hoff factor is 1.