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.

Answer

`3.225`

Explanation

`k_(f) = (8.314 xx(278.4)^(2) xx 78)/(1000 xx 10042) = 5`
`m = (0.02 xx 1000)/(0.98 xx 78) = 0.2614`
`DeltaT_(f) = i xx k_(f) xx m`
`i=1 -(alpha)/(2) = 0.7613`
`alpha = 0.4713`
`{:(2A,hArr,A_(2)),(C,,),(C-Calpha,,Calpha//2):}`
`k = (0.4713)/(2 xx 0.2614(1-0.4713)^(2)) = 3.225`