The freezing point of a solution containing of `0.2g` of acetic acid in `20.0g` of benzene is lowered `0.45^(@)C`. Calculate.
(i) the molar mass of acetic acid from this data
(ii) Van't Hoff factor
[For benzene, `K_(f)=5.12K kg "mol"^(-1)`]
What conclusion can you draw from the value of Van't Hoff factor obtained ?

Answer

Let the observed molecular mass of acitic acid.
moles `=(1000xxK_(f)xxW_(B))/(W_(A)xxDeltaT)=(1000xx5.12xx0.2)/(20xx0.45)`
moles `=113.78`
Normal molecular mass of acitic acid `=60`
`2CH_(3)COOHhArr(CH_(3)COOH)_(2)`
`{:(1,0),(1-alpha,alpha//2):}`
Van't Hoff factor `=("Normal molar mass")/("Observed molar mass")`
`i=(1-alpha+alpha//2)/(1) rArr(60)/(113.78)=1-(alpha)/(2)`
`alpha=0.945=94.5%`
Conclusion is that molecules get association whenever it mixed in acitic acid.