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 ?

Explanation

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%.