Question

Molecules of benzoic acid ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$ dimerises in benzene.$‘\mathrm{W}’\mathrm{g}$ of the acid dissolved in $30\mathrm{g}$ of benzene shows a depression in freezing point equal to $2\mathrm{K}$. If the percentage association of the acid to form a dimer in solution is$80$, then w is:

(Given that ${\mathrm{K}}_{\mathrm{f}}=5\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$, Molar mass of benzoic acid$=122\mathrm{g}{\mathrm{mol}}^{-1}$)

Answer 1

Step 1: Analyze the reaction

$\underset{\mathrm{Wg}}{2\left({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right)}\underset{30\mathrm{g}}{\overset{{\mathrm{C}}_6{\mathrm{H}}_6}{\to }}\underset{\mathrm{Dimer}}{{\left({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right)}_2}$

Step 2: Analyze the formula for depression in freezing point

${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{T}=\mathrm{i}\times {\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}$

• Here, ${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{T}$ is Depression in freezing point
• $\mathrm{i}$ is Van't hoff factor
• $\mathrm{m}$ is molality
• ${\mathrm{k}}_{\mathrm{f}}$ is cryoscopic constant

Step 3: Substitute the given values.

• ${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{T}=\mathrm{i}\times {\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}$
• Here, Van't hoff factor $\mathrm{i}=1-0.8+0.4=0.6$
• Now, $2=0.6\times 5\times \frac{\mathrm{W}\times 1000}{122\times 30}$
• $\mathrm{W}=2.44\mathrm{g}$

Thus, the value of $\mathrm{W}$ is $2.44\mathrm{g}$