Question

The molal freezing point constant of water is $1.86{{\text{ }}^ \circ }{\text{C kg mol}}{{\text{e}}^{ - 1}}$. The freezing point of $0.1{\text{ m NaCl}}$ solution is
A) $- {1.86^ \circ }{\text{C}}$
B) $- {0.372^ \circ }{\text{C}}$
C) $- {0.186^ \circ }{\text{C}}$
D) ${0.372^ \circ }{\text{C}}$

Answer 1

To solve this we must know that the freezing point of a pure solvent decreases when a non-volatile solute is added to it. This decrease in the freezing point is known as the depression in freezing point. Thus, when sodium chloride is added to water the freezing point of water decreases.

Complete solution: We know that the freezing point of a pure solvent decreases when a non-volatile solute is added to it. This decrease in the freezing point is known as the depression in freezing point. Thus, when sodium chloride is added to water the freezing point of water decreases.
We know that the equation for the depression in freezing point of a solution is,
$\Delta {T_f} = {K_f} \times m \times i$
Where, $\Delta {T_f}$ is the depression in freezing point,
${K_f}$ is the molal freezing point constant,
$m$ is the molality of the solution,
$i$ is the van’t Hoff factor.
${\text{NaCl}}$ dissociates into one ${\text{N}}{{\text{a}}^ + }$ ion and one ${\text{C}}{{\text{l}}^ - }$ ion i.e. total two ions. Thus, the van’t Hoff factor for ${\text{NaCl}}$ is 2.
We are given that the molal freezing point constant of water is $1.86{\text{ K kg mol}}{{\text{e}}^{ - 1}}$, the molality of the solution is $0.1{\text{ m}}$. Thus,
$\Delta {T_f} = 1.86{{\text{ }}^ \circ }{\text{C kg mol}}{{\text{e}}^{ - 1}} \times 0.1{\text{ m}} \times 2$
$\Delta {T_f} = {0.372^ \circ }{\text{C}}$
Thus, the depression in the freezing point of $0.1{\text{ m NaCl}}$ solution is ${0.372^ \circ }{\text{C}}$.
We know that the freezing point of water is ${0^ \circ }{\text{C}}$. Thus,
Freezing point $= {\left( {0 - 0.372} \right)^ \circ }{\text{C}}$
Freezing point $= - {0.372^ \circ }{\text{C}}$
Thus, the freezing point of $0.1{\text{ m NaCl}}$ solution is $- {0.372^ \circ }{\text{C}}$.

Thus, the correct option is (B) $- {0.372^ \circ }{\text{C}}$.

Note: The freezing point of a pure solvent decreases when any non-volatile solute is added to it. This is known as depression in freezing point. The freezing point of a pure solvent is always higher than that of its solution.