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The freezing point of solution containin...

The freezing point of solution containing `0.2 g` of acetic acid in `20.0 g` of benzene is lowered by `0.45^(@)C`. Calculate the degree of association of acetic acid in benzene.
`(K_(f)=5.12 K^(@) mol^(-1) kg^(-1))`

Text Solution

Verified by Experts

The correct Answer is:
0.946
Setp I. Calculation of observed molar mass of acetic acid
`M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A)), W_(A)=0.2 g, W_(A)=0.02 kg, DeltaT_(f)=0.45^(@)C=0.45 K`
`K_(f)=5.12" K kg mol"^(-1), M_(B)=((5.12" K kg mol"^(-1))xx(0.2 g))/((0.45 K)xx(0.2 kg ))=113.8" g mol"^(-1)`
Step II. Calculation of Van't Hoff factor (i)
`i=("Normal molar mass")/("Observed molar mass")=(60" g mol"^(-1))/((113.8" g mol"^(-1)))=0.527`
`"Step III. Calculation of degree of association "(alpha)`
`"For association "alpha=(i-1)/(1/n-1)=(0.527-1)/(1/2-1)=((-0.473))/((-0.50))=0.946= 0.946xx100=94.6%`
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