2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant (K_(f)) of benzene is "4.9 K kg mol"^(-1). What is the percentage association of the acid if it forms dimer in the solution?

2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depressio

1.	2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant (K_(f)) of benzene is "4.9 K kg mol"^(-1). What is the percentage association of the acid if it forms dimer in the solution?
Answer» Solution :Mass of solute (benzoic acid), `w_(2)=2.0 G",Mass of solvent (benzene), "w_(1)=25.0g` Observed `DeltaT_(f)=1.62K","K_(f)="4.9 K kg mol"^(-1)` `therefore"Observed molar mass of benzoic acid"` `M_(2)=(1000xxK_(f)xxw_(2))/(DeltaT_(f)xxw_(1))=("1000 g kg"^(-1)xx"4.9 K kg mol"^(-1)xx"2.0 g")/("1.62 K"xx"25.0 g")="242 g mol"^(-1)` Calcualted molar mass of benzoic acid `(C_(6)H_(5)COOH)=72+5+12+32+1="122 g mol"^(-1)` `"van't Hoff factor,i"=("Calculated mol. mass")/("Observed mol. mass")=(122)/(242)=0.504` If `alpha` is the degree of association of benzoic acid, then we have `{:(,2C_(6)H_(5)COOH,hArr,(C_(6)H_(5)COOH)_(2)),("Initial moles","1",,"0"),("After association",""1-alpha,,""alpha//2):}` `therefore"Total number of moles after association "=1-alpha+(alpha)/(2)=1-(alpha)/(2) therefore i=(1-alpha//2)/(1)=0.504 or 1-(alpha)/(2)=0.504` `alpha=(1-0.504)xx2=0.496xx2=0.992"or% AGE association = 99.2%".`