Question

Phenol associates in benzene to certain extent to form a dimer. A solution containing $20 \times {10^{ - 3}}{\text{ kg}}$ of phenol in $1.0{\text{ kg}}$ of benzene has its freezing point depressed by $0.69{\text{ K}}$. Calculate the fraction of phenol that has dimerized. (${K_f}$ for benzene is $5.12{\text{ K kg mo}}{{\text{l}}^{ - 1}}$).

Answer 1

Hint: Phenol is dimerized means that two molecules of phenol combine and form a dimer. The decrease in the freezing point of pure solvent when a non-volatile solute is added to it is known as depression in freezing point.

Formulae Used:
${\text{Molality}}\left( {\text{m}} \right){\text{ = }}\dfrac{{{\text{Number of moles}}\left( {{\text{mol}}} \right)}}{{{\text{Mass of solvent}}\left( {{\text{kg}}} \right)}}$
$\Delta {T_f} = {K_f} \times m \times i$

Complete step by step solution:
We have the reaction of dimerization of phenol,
${\text{2}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} \rightleftharpoons {\left( {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}} \right)_{\text{2}}}$
In the reaction, two molecules of phenol combine and form a dimer.
Thus, the equilibrium table is as follows:
${\text{2}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} \rightleftharpoons {\left( {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}} \right)_{\text{2}}}$
$\;\;\;\;\;\;\;1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0 \;\;\;\;\;\;\;\;\;\; $ …… Initial moles
     $1 - \alpha \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ $\alpha {\text{/2}}$ $\;\;\;\;\;\;$ …… Moles at equilibrium
Thus, the total moles of the reactant and product at equilibrium are as follows:
Total moles $ = 1 - \alpha + \dfrac{\alpha }{2}$
Total moles $ = 1 - \dfrac{\alpha }{2}$
Where, $\alpha $ is the degree of dissociation.
The number of individual ions any ionic compound dissociates into is known as the van’t Hoff factor. Thus,
$i = 1 - \dfrac{\alpha }{2}$
Where $i$ is the van’t Hoff factor.
The molality of the solution is the number of moles of solute per kilogram of solvent. Thus,
${\text{Molality}}\left( {\text{m}} \right){\text{ = }}\dfrac{{{\text{Number of moles}}\left( {{\text{mol}}} \right)}}{{{\text{Mass of solvent}}\left( {{\text{kg}}} \right)}}$
Thus,
${\text{Molality}} = \dfrac{{{\text{Number of moles of phenol}}}}{{{\text{Mass of benzene}}}}$
${\text{Molality}} = \dfrac{{{\text{Mass of phenol/Molar mass of phenol}}}}{{{\text{Mass of benzene}}}}$
Substitute $20 \times {10^{ - 3}}{\text{ kg}} = 20{\text{ g}}$ for the mass of phenol, $94{\text{ g mo}}{{\text{l}}^{ - 1}}$ for the molar mass of phenol, $1.0{\text{ kg}}$ for the mass of benzene. Thus,
$\Rightarrow {\text{Molality}} = \dfrac{{{\text{20 g/}}94{\text{ g mo}}{{\text{l}}^{ - 1}}}}{{1.0{\text{ kg}}}}$
$\Rightarrow {\text{Molality}} = 0.2127{\text{ m}}$
Thus, the molality of the solution is $0.2127{\text{ m}}$.
The formula for the depression in freezing point is,
$\Delta {T_f} = {K_f} \times m \times i$
Where
$\Delta {T_f}$ is the depression in freezing point,
${K_f}$ is the freezing point depression constant,
$m$ is the molality of the solution,
$i$ is the van’t Hoff factor
Substitute $0.69{\text{ K}}$ for the depression in freezing point, $5.12{\text{ K kg mo}}{{\text{l}}^{ - 1}}$ for the freezing point depression constant, $0.2127{\text{ m}}$ for the molality of the solution, $1 - \dfrac{\alpha }{2}$ for the van’t Hoff factor. Thus,
$\Rightarrow 0.69{\text{ K}} = 5.12{\text{ K kg mo}}{{\text{l}}^{ - 1}} \times 0.2127{\text{ m}} \times i$
$\Rightarrow i = \dfrac{{0.69{\text{ K}}}}{{5.12{\text{ K kg mo}}{{\text{l}}^{ - 1}} \times 0.2127{\text{ m}}}}$
$\Rightarrow i = 0.633$
But $i = 1 - \dfrac{\alpha }{2}$. Thus,
$\Rightarrow 1 - \dfrac{\alpha }{2} = 0.633$
$\Rightarrow 2 - \alpha = 1.266$
$\Rightarrow \alpha = 2 - 1.266$
$\Rightarrow \alpha = 0.734$

Thus, the fraction of phenol that has dimerized is $0.734 \times 100 = 73.4\% $.

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.