Question

Molecules of benzoic acid $\left( {{C_6}{H_5}COOH} \right)$ dimerse in benzene. ‘w’ g of the acid dissolved in 30g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form a dimer in the solution is 80, then w is:
(Given that ${K_f}$=$5\,K\,kg\,mo{l^{ - 1}}$, molar mass of benzoic acid=$122\,g\,mo{l^{ - 1}}$)
A.1.28 g
B.2.4 g
C.1.0 g
D.1.5 g

Answer 1

Hint:We can calculate the weight of the acid using the freezing point depression, Van’t Hoff factor, freezing point depression constant, mass of benzene (solvent), and the molar mass of benzoic acid.
Formula used: The freezing point depression is proportional to the concentration of the solute particles and is given by the equation,
$\Delta {T_f} = im{k_f}$
Where,
$\Delta {T_f} = $Freezing point depression
\[m = \]Molal concentration
${k_f} = $Freezing point depression constant (depends on the solvent used).
$i = $Van’t Hoff factor

Complete step by step answer:
Given data contains,
Mass of benzene is 30g.
Depression in the freezing point is 2K.
Molar mass of benzoic acid is $122\,g\,mo{l^{ - 1}}$.
Freezing point depression constant is $5\,K\,kg\,mo{l^{ - 1}}$.
Percentage association is 80%.
We can write the chemical equation as,

We have to calculate the van't Hoff factor using percentage dissociation,

$x = 80\% $
$x = 0.8$
The van't Hoff factor is calculated as,
$i = 1 - x + \dfrac{x}{2}$
$i = 1 - \dfrac{x}{2}$
$i = 1 - \dfrac{{0.8}}{2}$
$i = 0.6$
The van't Hoff factor is $0.6$.
Let us now calculate the mass of the acid.
We can use the formula of freezing point depression to calculate the mass of the acid. We can write the formula as,
$\Delta {T_f} = im{k_f}$
Substituting the values of freezing point depression, van't Hoff factor, molality, freezing point depression constant, we can calculate the mass of the acid as,
$\Delta {T_f} = im{k_f}$
$2 = 0.6 \times 5 \times \dfrac{{w \times 1000}}{{122 \times 30}}$
$w = 2.44\,g$
The mass of the acid dissolved is $2.44\,g$.
Therefore, the option (B) is correct.


Note:
We also remember that the freezing point depression is a colligative property. We can also calculate the freezing point depression constant and molality using the expression of freezing point depression of solution.
We can calculate the Van’t Hoff factor using the equation,
${\text{i = }}\dfrac{{{\text{measured}}\,{\text{colligative}}\,{\text{property}}}}{{{\text{expected}}\,{\text{value}}\,{\text{for}}\,{\text{a}}\,{\text{nonelectrolyte}}}}$
For solutions that are nonelectrolytes, like urea and sucrose, the Van’t Hoff factor is one. For solutions of salts and other electrolytes, the value of i is greater than one. For dilute solution such as (${\text{0}}{\text{.01m}}$ or less), the Van’t Hoff factor will be equal to the number of ions formed by each formula unit of the compound that dissolves.
Example: We can calculate Van’t Hoff factor of sulfuric acid as below,
Given,
Mass percentage of the solution =${\text{2}}{\text{.00% }}$
Freezing point of the solution =${\text{ - 0}}{\text{.79}}{{\text{6}}^{\text{o}}}{\text{C}}$
The change in the freezing point is calculated as,
$\Delta {T_f} = {T_o} - {T_f}$
$\Delta {T_f} = {0.00^o}C - ( - {0.796^o}C)$
$\Delta {T_f} = {0.796^o}C$
The moles of sulphuric acid are calculated from its molar mass.
Moles of sulphuric acid=$2.00g \times \dfrac{{1\,mol}}{{98.07g}} = 0.0203moles$
Moles of sulphuric acid=$0.0203moles$
The mass of the solvent is calculated as,
Mass of the solvent=$98ml \times \dfrac{{1\,g}}{{ml}} = 98g$
Mass of the solvent in kilograms=$98g \times \dfrac{{1\,kg}}{{1000g}} = 0.098kg$
The molality of the solution is calculated as,
Molality=$\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{sulphuric}}\,{\text{acid}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{solvent}}}}$
Molality=$\dfrac{{0.0203moles}}{{0.098kg}}$
Molality=$0.2071m$
The Van’t Hoff factor is calculated as,
$i = \dfrac{{\Delta {T_f}}}{{m{K_f}}}$
$i = \dfrac{{{{0.796}^o}C}}{{(0.2071m)({{1.86}^o}C/m)}}$
$i = \dfrac{{{{0.796}^o}C}}{{{{0.3852}^o}C}}$

The Van’t Hoff factor of sulphuric acid is $2.06 \approx 2.1.$