Question

$2.0$ g of benzoic acid dissolved in $25.0$g of benzene shows a depression in freezing point equal to $1.62$K. Molar depression constant $\left(K_f\right)$ of benzene is $4.9K.kg.mo{l}^{-1}$. What is the percentage association of the acid?

Answer 1

Depression of freezing point, ${\Delta T}_f=K_f\times \frac{wt.ofbenzoicacid\times 1000}{Mol.wtofbenzoicacid(wt.ofbenzene)}$

$\Rightarrow {\Delta T}_f=\frac{2\times 1000\times 4.9}{Eff.mol.wtofbenzoicacid\times 25}$

$\Rightarrow$ Effective mol. wt of benzoic acid $=241.98$ gm/mol

$2C_6{H}_{5}COOH\rightleftharpoons {\left(C_6{H}_{5}COOH\right)}_2$

if ${}^{\prime }{x}^{\prime }$ is degree of dissociation,

total number of moles of particles at equilibrium,

$i=1-x+x/2=1-x/2$

$i=\frac{Normalmolecularweight}{Effectivemolecularweight}=\frac{122}{241.98}$

$\therefore$ $\frac{122}{241.98}=1-x/2$

$\Rightarrow x=\frac{239.96}{241.98}=0.992=99.2$%

$\therefore$ Therefore degree of dissociation of benzoic acid is benzene is $99.2$%