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

`DeltaT_(f)=K_(t)xx mxx (i)" "....(1)`
for acetic acid : `C " " 0`
`(1-alpha) " " (alpha//2)`
`i=1-alpha+alpha/2=(1-alpha//2)`
`rArr i=(1-alpha//2)" ....(ii)`
`K_(f)=(RT^(2))/(1000xxi_(f))=(8.3145xx(278.4)^(2))/(1000xx(10.042xx10^(3))/(78))`
`=5.0` molality
`DeltaT_(f)=278.4-277.4=1`
Given, mol fraction of acetic acid `=0.02=(n_(1))/((n_(1)+n_(2)))`
Mole fraction of benzene `=(n_(2))/((n_(1)+n_(2)))=0.98`
`(n_(1))/(n_(2))=(0.02)/(0.98)`
Molality `=(n_(1))/(w_(2))xx1000=(n_(1)xx1000)/(n_(2)xxm_(2))`
`=(0.02)/(0.98)xx(1000)/(78)`
`0.262 m`.
From equation (i) and (ii)
`1=50xx0.0262xx(1-alpha//2)`
`:.alpha=0.48`
`K_(c)=((calpha//2))/([c(1-alpha)]^(2))=(alpha)/(2xxcxx(1-alpha)^(2))=(0.48)/(2xx(0.262)^(2)xx(1-0.48)^(2))`
`3.39`