Question

2 g benzoic acid $\left(C_6{H}_{5}COOH\right)$ dissolved in 25 g of benzene shoes a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg $mo{l}^{-1}$. What is the percentage association of acid if it forms dimer in solution?

Answer 1

Here solute is bengoic acid and soluent is bengelnt

$W$solute $=2g$

$M$solute $\left(6H_6\right)=6\times 12+6=78g/mole$

$K_f=4.9Kkg/mole$

$\Delta T_f=1.62K$

$2C_6{H}_{5}COOH\rightleftharpoons (6H_{5}COOH{)}_2$

If ${}^{\prime }{x}^{\prime }$ is degree of association then we would have $(1-x)$ mol of benxoic acid left undissociation.

Now total no. of moles of particles at equilibrium

$=1-x+\frac{x}{2}=1-x---\left(1\right)$

$=i$

$i=\frac{NormalMolecularmass}{NormalMolecularmass}$

$=\frac{122}{241.98}----\left(2\right)$

$M\left(abnormal\right)=\frac{4.9\times 2\times 1000}{1.62\times 25}=241.98g/mol$

From $\left(1\right)$ & $\left(2\right)\frac{122}{241.98}=-\frac{x}{2}$

$\frac{x}{2}=\frac{241.98-122}{241.98}=99.2\%$

Degree of association of benzoic zcid $=99.2\%$