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

Let us first calculate observed molar mass ,
`M_(B) = (k_(f) xx w_(B) xx 1000)/(Delta T_(f) xx w_(A))`
`w_(B) = 2 g , w_(A) = 25 g, Delta T_(f) = 1.62 K`
`k_(f) = 4.9 K Kg mol^(-1)`
`M_(B) = (4.9 xx 2 xx 1000)/(1.62xx 25) = 241.98 g mol^(-1)`
Observed molar mass ` = 241.98 mol^(-1)`
Normal molar mass of `C_(6)H_(5)COOH`
` = 7 xx 12+6 xx 1 + 2 xx 16 = 122 g mol^(-1)`
Benzoic acid assoicates as :
`2C_(6) H_(5) COOH hArr (C_(6)H_(5)COOH_(2))`
If `alpha` is the degree of association .
`{:(,"Initial moles","1","0"),(,"Moles after association",1-alpha,alpha//2):}`
Total moles after association `= 1 -alpha + alpha//2 = 1 - alpha//2`
`i=("Normal molar mass")/("Observed molar mass")= (22)/(241.98) = 0.504`
Now `" " i=(1-(alpha)/(2))/(1) =0.504`
` - (alpha)/(2) = 0.504`
` - (alpha)/(2) = - 0.496`
` alpha = 0.496 xx 2 = 0.992`
Thus, degree of association of benzoic acid in benzene = 99.2%