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

Let acetic acid =A
Benzene =B
Assume `alpha` part of A forms dimer
`{:(2A ,hArr, A_(2)),(1,0,"Intial moles"),(1-alpha, alpha//2 ,"moles after at eqm"):}`
`:.I=((1-alpha)+alpha//2)/(1)=1-alpha//2`
Mol. Fraction of `A=X_(A)=0.02`
Mol. fraction of `B=X_(B)=0.98`
Molality of A in B `=(X_(A))/(m_(1))xx(1000)/(X_(B))=(0.02)/(78)xx(1000)/(0.98)= 0.262 "mol" kg^(-1) (m_(1)="mol"` wt of solvent)
Since `DeltaT_(t)=K_(1)xxlxx"molality"`
`278.4-277.4=5xxixx0.262`
or `1=5xxixx0.262`
`i=(1)/(5)xx0.262=0.763`
`1-alpha//2=0.763 rArr alpha=0.48`
Hence the molality of A after dimer is formed `(1-alpha)xx` initial molality
`=0.52xx0.262`
Molality of `A_(2)` after dimer formed `=(alpha)/(@)xx"molality" =(0.48)/(2)xx 0.262`
`=0.24xx0.26=0.06288`
The equilibrium constant
`K_(eq)=([A_(2)])/([A]^(2))=(0.06288)/((0.13624)^(2))=3.39`