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))

The freezing point of solution containing 0.2 g of acetic acid in 20.0

1.	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))
Answer» Solution :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%`