Question

2 g of benzonic acid dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62K. Molal depression constant for benzene is 4.9kg/mol .what is the percentage association of acid if it forms dimer in solution

Answer 1

Dear student,

Since : $∆T_f=i\times K_f\times molality$ and molality = n / W (in kg) again , n = w / M = 2 / 122 = 0.016 moles
Then the molality = 0.016 / 25 $\times$ 10^-3 = 0.64 m
Now on putting the values in the above relation : 1.62 = i $\times$ 4.9 $\times$ 0.64 , Then i = 0.5
Again from the relation between i and degree og dissociation : i = 1- $\alpha$ (1 - 1 /n) where n = 2 ( as benzoic acid dimerises)
then $\alpha$ = 1 , and % of dissociation = 100 %
Regards