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 ?

2 g of benzoic acid (C_(6)H_(5)COOH)dissolved in 25 g of benzene shows

1.	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 ?
Answer» Solution :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 %.