The freezing point of `0.02` mole fraction acetic acid in benzene is `277.4 K`. Acetic acid exists partly as dimer. Calculate the equilibrium constant for dimerization. The freezing point of benzene is `278.4 K` and the heat the fusion of benzene is `10.042 kJ mol^(-1)`. Assume molarity and molality same.

Explanation

For benzene ,`K'_(f)=(RT^(2))/(1000l_(f)(cal g^(-1)))`
`=(8.34 xx (278.4)^(2))/(1000xx(10.042 xx 10^(3))/(78)) = 5.0 K` molality
Also,
For acetic acid in benzene

Before association
After association
`:.K_(C)=2/(C^(2)(a-alpha)^(2))`
where `alpha` is the degree of association. ...(i)
Also `DeltaT = K'_(f)xx "molality"xx(1-alpha/2)` ........(ii)
`( :'` Total particles at equilibrium `= 1-alpha+alpha/2=1-alpha/2)`
Given mole fraction of acetic acid `=0.02 =n_(2)(n_(2)+n_(2))`
`:.` Mole fraction of benzene`=0.98=n_(2)/(n_(2)+n_(2))`
`n_(2)/n_(1) =0.02/0.98`
`:.` Molality`=n_(2)/W_(1)xx1000`
`=(n_(2) xx 1000)/(n_(1) xx Mw_(1)) =(0.02 xx 1000)/(0.98 xx 78) = 0.262m`
`:.` From Eq.(ii), `1=5xx0.262 xx(1-alpha/2)`
`:.alpha=0.48`
From Eq.(i),assuming molarity = molarity
`K_(C) =(0.262xx0.48)/(2xx(0.262)^(2)xx(1-0.48)^(2)) =3.39`