What is the freezing of 0.4 molal solution of acetic acid in benzene in which it dimerises to the extent of Freezing point of benzene is 278.K and its molar heat of fusion is 10-042KJ `mol^(-1)`

Explanation

Calculation of molal depression constant `(k_(p))` for benzene
`K_(f)=(R(T_(f)^(@))^(2)M)/((DeltaH_("fusion")))`
`R=8.314 JK^(-1) mol^(-1), T_(f)^(@)=278.4K`
`M_(Benzene)=778 g mol^(-1)=0.078 kg mol^(-1),DeltaH_("Fasion")=10-.04xx10^(3)J mol^(-1)`
`K_(f)=((8.314 JK^(-1)mol^(-1))xx(278.4 K)^(2)xx(787g mol^(-1)))/((10.04xx10^(3)jmol^(-1)))`
=5.0 K kg `mol^(-1)`
Calculation of Van't Hoff factor (i)
In benzene solventn, acetic acid dimerises as:
`2CH_(3)COOHhArr((CH_(3)COOH)_(2)`
`alpha=85%=0.85, n=1//2 `
`alpha=(i-1)/(1//n-1),0.85=(i-1)/((1//2-1))`
`i=0.85 (-1//2)+1==0.575`
Calculation of freezing point of solution
`i=0.575, K_(f)=5.0 K kg mol^(-1), m=0.4m 0.4 kg , mol^(-1)`
`Delta_(f)=i K_(f)m=(0.575)xx(5.0 K kg mol^(-1))xx(0.4 kg mol^(-1))`
Freezing point solution `(T_(f))=(278.4-1.15)=277.25` K