Question

Which one of these has the highest freezing point?
(i) $1M$ glucose
(ii) $1M$ NaCl
(iii) $1M$ $CaC{l_2}$

Answer 1

Colligative properties are those properties of solution which depend directly on the mass of solute or amount of solute which is present in the solution. One such property of solutions is depression in freezing point. In order to compare the freezing points of these given compounds, we should understand the concept of the van't hoff factor.

Complete Step By Step Answer:
We will first see the formula for depression in freezing point and it is given as:
$\Delta {T_f} = i{K_f}m$
In the above equation, $\Delta {T_f}$ is the depression in freezing point of the solution, $i$ is the van’t hoff factor of the solute, ${K_f}$ is the constant and m is the molality of the solution.
Here, the molality of all the given solutions are same, so the depression in freezing point depends on the van’t hoff factor.
Van’t hoff factor is the number of ions in which the solute breaks down upon dissolving in the solvent. Therefore, greater the Van’t hoff factor, greater will be the depression in freezing point and lesser will be the freezing point of the solution.
We will now calculate the van’t hoff factor for each of the given solute as follows:
(i) $1M$ glucose
Glucose does not dissociate into ions, so its van’t hoff factor will be one.
(ii) $1M$ NaCl
$NaCl \rightleftharpoons N{a^ + } + C{l^ - }$
$i = 2$
(iii) $1M$ $CaC{l_2}$
$CaC{l_2} \rightleftharpoons C{a^{2 + }} + 2C{l^ - }$
$i = 3$
Therefore, the order of depression in freezing point is:
$CaC{l_2} > NaCl > glu\cos e$
Hence, glucose will have the highest freezing point.
Therefore, the correct option is: (i) $1M$ glucose.

Note:
We should remember that the van’t hoff factor determines the degree of dissociation, association of a non-volatile solute when it is dissolved in a solvent. There are some more colligative properties such as elevation in boiling point, osmotic pressure and relative lowering of vapour pressure.