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 $4.9Kkgmo{l}^{-1}$. What is the percentage association of acid if it forms dimer in solution?

Answer 1

Given:
Mass of solute $\left(\text{benzoic acid}\right),\left(m_2\right)=2g$,
Molal depression constant for benzene $\left(K_f\right)=4.9Kkgmo{l}^{-1}$,
Mass of solvent $\left(benzene\right),\left(m_1\right)=25g$,
Depression in freezing point $\left(\Delta T_f\right)=1.62K$
Molar mass of benzoic acid

We know that $\Delta T_f=K_f\times \text{molality(m)}$
We also know that molality$=\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{mass of solvent(kg)}}$

$\Delta T_f=K_f\times \frac{\text{mass of solute}\left(m_2\right)}{\text{Molar mass of solute}\left(M_2\right)\times \text{mass of solvent(m\_1)}}$
So, $M_2=K_f\times \frac{\text{mass of solute}\left(m_2\right)}{\Delta T_f\times \text{mass of solvent}\left(m_1\right)}$

$M_2=\frac{4.9Kkgmo{l}^{-1}\times 2g}{1.62K\times 25\times {10}^{-3}kg}$
$=241.98gmo{l}^{-1}$
Thus, experimental molar mass of benzoic acid in benzene is $=241.98gmo{l}^{-1}$
Degree of association of benzoic acid

Now consider the following equilibrium for the acid:
$2C_6{H}_{5}COOH\rightleftharpoons (C_6{H}_{5}COOH{)}_2$
$\begin{array}{ccc}\text{At t}=0& 1& 0\\ \text{At t}=t_{eq}& 1-x& \frac{x}{2}\end{array}$
Therefore, total number of moles of particles at equilibrium is:
$1-x+\frac{x}{2}=1-\frac{x}{2}$

Thus, total number of moles of particles at equilibrium equals van't Hoff factor $\left(i\right)$
But we know that,

$i=\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$

$=\frac{122gmo{l}^{-1}}{241.98gmo{l}^{-1}}$

So,$1-\frac{x}{2}=\frac{122}{241.98}$

or,$\frac{x}{2}=1-\frac{122}{241.98}=1-0.504=0.496$
or,$x=2\times 0.496=0.992$

Therefore, degree of association of benzoic acid in benzene $=99.2\%$.