Two grams of benzoic acid `(C_(6)H_(5)COOH)` dissolved in `25.0 g` of benzene shows a depression in freezing point equal to `1.62 K`. Molal depression constant for benzene is `4.9 K kg^(-1)"mol^-1`. What is the percentage association of acid if it forms dimer in solution?

Explanation

The given quantities are `w_(2)=2g,K_(f)=4.9" K kg "mol^(-1),w_(1)=25g`,
`DeltaT_(f)=1.62K`
Substituting these values in equation we get :
`M_(2)=(4.9" K kg "mol^(-1)xx2gxx1000g" g "kg^(-1))/(25gxx1.62K)=241.98" g "mol^(-1)`
Thus, exeriment molar mass of benzoic acid in benzene is
`=241.98" g "mol^(-1)`
Now consider the following equilibrium for the acid:
`2C_(6)H_(5)COOHhArr(C_(6)H_(5)COOH_(2))`
If x represents the degree of association of the solute then we would have (1-x) mol of benzoic acid left in unassociated form and
correspondingly `(x)/(2)` as associated moles of particles at equilibrium is:
`1-x+(x)/(2)=1-(x)/(2)`
Thus, total number of moles of particles at equilibrium equals van't Hoff factor i.
But `i=("Normal molar mass")/("Abnormal molar mass")`
`=(122" g "mol^(-))/(241.98" g "mol^(-1))`
or `(x)/(2)=1-(122)/(241.98)=1-0.504=0.496`
or `x=2xx0.496=0.992`
Therefore, degree of association of benzoic acid in benzene is 99.2%