2 g of benzoic acid `(C_(6)H_(5)COOH)` dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molar depression constant for benzene is 4.9 K kg `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)=25 g, " " Delta T_(f)=1.62 K`
Substituting these values in equation,
`M_(2)=(K_(f)xx w_(2)xx1000)/(Delta T_(f)xx w_(1))`
`= (4.9"K kg mol"^(-1)xx 2g xx 1000 g kg^(-1))/(25 g xx 1.62 K)`
`= 241.98 g mol^(-1)`
Thus, experimental 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)COOH hArr (C_(6)H_(6)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 asociated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is :
`1-x+(x)/(2)=1-(x)/(2)`
Thus, total number of moles of particles ast equilibrium equals vant.s Hoff factor (i).
But `i=("Normal molar mass")/("Abnormal molar mass")`
`=(122 g mol^(-1))/(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 %.