Question

The molal freezing point constant for water is $1.86\,{\text{K kg mo}}{{\text{l}}^{ - 1}}$. The freezing point of ${\text{0}}{\text{.1}}\,{\text{m}}$ ${\text{NaCl}}$ solution is
A. $- 1.{\text{86}}{\,^{\text{o}}}{\text{C}}$
B. $- 0.372{\,^{\text{o}}}{\text{C}}$
C. $- 0.186{\,^{\text{o}}}{\text{C}}$
D. $0.372{\,^{\text{o}}}{\text{C}}$

Answer 1

The depression in freezing point is the product of the freezing point depression constant, Van't hoff factor and molality. Van't hoff factor is determined as the number of ions produced by the ionic compound in the solution.

Formula used: $\Delta {T_f}\, = i\,{K_f}.\,m$

Complete answer:
The formula of freezing point depression is as follows:
$\Delta {T_f}\, = i\,{K_f}.\,m$
Where,
$\Delta {T_f}$ is the depression in freezing point.
${K_f}$ is the freezing point depression constant.
$i\,$ is the van't Hoff factor.
$m$ is the molal concentration
Sodium chloride is an ionic compound which dissociates in water as follows:
${\text{NaCl}}\mathop \to \limits^{{{\text{H}}_{\text{2}}}{\text{O}}} \,{\text{N}}{{\text{a}}^{\text{ + }}}\, + \,{\text{C}}{{\text{l}}^ - }$
Sodium chloride produces two ions so the value of van't Hoff factor is $2$.
We will use the formula of freezing point depression to determine the depression in freezing point as follows:
Substitute $2$ for van't Hoff factor, $1.86\,{\text{K kg mo}}{{\text{l}}^{ - 1}}$ for freezing point depression constant and $0.1\,{\text{m}}$ for molal concentration.
$\Delta {T_f}\, = 2\, \times 1.86\,{\text{K kg mo}}{{\text{l}}^{ - 1}} \times 0.1\,{\text{mol kg}}^{-1}$
$\Delta {T_f}\, = 0.372\,{\text{K}}$
So, the depression in freezing point is $0.372\,{\text{K}}$.
The relation in freezing point temperature of the solution, solvent and freezing point depression is as follows:
${{\text{T}}_{{\text{solution}}}}\,{\text{ = }}\,{{\text{T}}_{{\text{solvent}}}}\, - \,\Delta {{\text{T}}_{\text{f}}}\,$
The freezing point temperature $0{\,^{\text{o}}}{\text{C}}$.
So, substitute $0.372{\,^{\text{o}}}{\text{C}}$ for $\Delta {{\text{T}}_{\text{f}}}\,$ and $0{\,^{\text{o}}}{\text{C}}$ for ${{\text{T}}_{{\text{solvent}}}}$.
${{\text{T}}_{{\text{solution}}}}\,{\text{ = }}\,0{\,^{\text{o}}}{\text{C}}\, - \,0.372{\,^{\text{o}}}{\text{C}}$
${{\text{T}}_{{\text{solution}}}}\,{\text{ = }}\, - \,0.372{\,^{\text{o}}}{\text{C}}$
So, the freezing point of {\text{0}}{\text{.1}}\,{\text{m }}$$${\text{NaCl}}$ solution is - ,0.372{,^{\text{o}}}{\text{C}}$$.

**Therefore, option (B)$- \,0.372{\,^{\text{o}}}{\text{C}}$, is correct.

Note:**
When a non-volatile solute is added to the pure solvent, the vapour pressure of the solution decreases. So, the freezing point of the solution decreases which is known as the depression in freezing point. The freezing point of the solution will be less than the freezing point of the solvent. The freezing point of the pure solvent at a temperature is defined as the freezing point constant. Van't Hoff factor represents the degree of dissociation, association or number of ions produced by a compound on dissolution.