Question

$2g$ of benzoic acid $({C_6}{H_5}COOH)$ dissolved in $25g$ of benzene shows a depression in freezing point equal to$1.62K$. Molal depression constant for benzene is $4.9{\text{ kg }}{{\text{m}}^{ - 1}}$. What is the percentage association of acid if it forms a dimer in solution?
A. $99.2\%$
B. $97.4\%$
C. $86.0\%$
D. $87.5\%$

Answer 1

Depression in freezing point: It is defined as the decrease in freezing point of a solvent on addition of non-volatile solute.
Elevation in boiling point: It is defined as an increase in boiling point of a solvent in addition to non-volatile solute.

Complete step by step answer:
First of all we will consider elevation in boiling point, depression in freezing point, osmotic pressure and relative lowering of vapour pressure.
Depression in freezing point: It is defined as the decrease in freezing point of a solvent in addition to non-volatile solute.
Elevation in boiling point: It is defined as an increase in boiling point of a solvent in addition to non-volatile solute.
Osmotic pressure: It is defined as the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.
Relative lowering in vapour pressure: When non- volatile solutes are added to a solution then there is a decrease in its vapour pressure, which is known as relative lowering in vapour pressure.
Now according to the question $2g$ of benzoic acid $({C_6}{H_5}COOH)$ dissolved in $25g$ of benzene shows a depression in freezing point equal to $1.62K$.
Weight of solute ${W_a} = 2g$
Weight of solution ${W_b} = 25g$, $\Delta {T_f}$ which is temperature difference is $1.62K$ and molal depression constant ${K_f} = 4.9K{m^{ - 1}}$
The formula for depression in freezing point is $\Delta {T_f} = {K_f} \times \dfrac{{{W_b}}}{{{M_b}}} \times \dfrac{{1000}}{{{W_a}}}$where ${M_b}$ is molar mass of solvent.
${M_b} = \dfrac{{4.9 \times 2 \times 1000}}{{25 \times 1.62}} = 241.98g/mol$
As we know that the benzoic acid forms dimer as $2{C_6}{H_5}COOH \to {({C_6}{H_5}COO)_2}$
Let $x$ be the degree of association then $(1 - x)$mole of benzoic acid will remain undissociated and $\dfrac{x}{2}$ moles of associated benzoic acid will be formed. So the total moles will be $1 - x + \dfrac{x}{2} = 1 - \dfrac{x}{2}$. This will be equal to $i$ known as van’t Hoff factor.
$i = \dfrac{{{\text{normal molecular mass}}}}{{{\text{abnormal molecular mass}}}}$ Normal molecular mass for benzoic acid is $12 \times 6 + 6 + 12 + 32 = 122$. And abnormal molecular mass we calculate as $241.98$. So, by putting these values we get $x$ as:
$1 - \dfrac{x}{2} = \dfrac{{122}}{{241.98}} \\\ x = 0.992 \\\$
Hence, the degree of percentage association is $99.2\%$.

So, the correct answer is Option A .

Note:
Van’t Hoff factor is the ratio of normal molecular mass (i.e. the molecular mass before adding non-volatile solute) to the abnormal molecular mass (i.e. the molecular mass after adding non-volatile solute).