2 g of benzoic acid (C_6H_5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal 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_6H_5COOH) dissolved in 25 g of benzene shows a

1.	2 g of benzoic acid (C_6H_5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal 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 data provided is listed below : `W_B = 2g, K_f = 4.9 K kg "mol"^(-1) , W_A = 25 g, DELTA T_f = 1.62 K` Applying the following relation : ` Delta T_f = m xx K_f = (W_B)/(M_B) xx (1000)/(W_A) xx K_f` Substituting the values, we get `1.62 = (2)/(M_B) xx 1000/25 xx 4.9` `M_B = (2000 xx 4.9)/(25 xx 1.62) = 241.98 g"mol"^(-1)` `underset(1 -x)underset(1)(2C_6H_5COOH) iff underset(x//2)underset(0)((C_6H_5COOH)_2)` Total number of particles at equilibrium = 1 - x + x/2 = 1 - x/2 `i = (1 - x//2)/(1) = 1 - x//2` i = Calculated molecular MASS / Observed molecular mass ` =(122)/(241.98)` Thus `(122)/(241.98) = 1 - x/2` `x/2 = 1 - (122)/(241.98) = (241.98 - 122)/(241.98) = (119.98)/(241.98) = 0.4958` ` x = 0.9916 = 99.16%` The degree of association of BENZOIC ACID in benzene is 99.16%.