Question

The value of ${K_f}$ for water is 1.86$kJmol^{-1}$, calculated from glucose solution. The value of ${K_f}$ for water calculated from $NaCl$ solution will be:
A. = $1.86$ $kJmol^{-1}$
B. < $1.86$ $kJmol^{-1}$
C. > $1.86$ $kJmol^{-1}$
D. Zero

Answer 1

The value of ${K_f}$ is related to that of freezing point depression. Freezing point depression of a solution can be described as the phenomenon of decrease in the freezing point of a pure solvent in addition to solute in the solvent. Freezing point depression is a colligative property that means it will depend upon the concentration of solute in the solvent, but independent of quantity and identity of solute.

Complete answer:
Let us first understand why the freezing point depression of water occurs.
When water freezes, water is converted into ice which has characteristic hydrogen bonding. The hydrogen bonding in ice leads to the formation of hexagonal shaped molecular networks. When a water-soluble solute is added to water, the solute starts dissolving in the water. This leads to the breaking of particle arrangement in the solvent. Because of this, to freeze water more energy will be needed to remove form water molecules to become ice. This results in a decrease in the freezing point of water.
The molality of the solution formed by solute and solvent is related to that of the freezing point depression of the solvent. The freezing point depression is directly proportional to that of the molality of that solution.
$\vartriangle {T_f} = {K_f} \times m$
The value of $\vartriangle {T_f}$ is freezing point depression of the solvent and m is the molality of the solution.
The value of ${K_f}$ is given, which is known as molal freezing point depression constant or simply freezing point depression constant. The value of freezing point depression constant can be considered as the change that is taking place in the freezing point of 1-molal solution when a non-volatile solute is dissolved in it.
The value of ${K_f}$ for water is $1.86$ $kJmol^{-1}$.
When glucose is dissolved in water and the freezing point depression is calculated for the solution, it will be $1.86$ $kJmol^{-1}$. The value of ${K_f}$ for water will remain $1.86$ $kJmol^{-1}$ on dissolving $NaCl$ or sodium chloride in the water. This value is unique for water and it will remain the same for any solution of water as it is a constant.

**Hence, option (A) is the correct option.

Note: **
The concept of vapour pressure can be used to explain the freezing point depression of a pure solvent. According to Roults’ law, the vapour pressure will also decrease when a solute is added to the solvent and it disrupts the particle arrangement in that solvent.
Freezing point depression of a solvent can also be calculated using the following formula.
$\vartriangle {T_f} = i \times {K_f} \times m$
$\vartriangle {T_f}$ is freezing point depression of the solvent
$i$ is Van’t Hoff factor
${K_f}$ is freezing point depression constant
m is molality of the solution