The freezing point depression of 0.1 M molal solution of acetic acid in benzene is 0.256 K. K_(f) for benzene is "5.12 K kg mol"^(-1). What conclusion can you draw about the molecular state of acetic acid in benzene?

The freezing point depression of 0.1 M molal solution of acetic acid i

1.	The freezing point depression of 0.1 M molal solution of acetic acid in benzene is 0.256 K. K_(f) for benzene is "5.12 K kg mol"^(-1). What conclusion can you draw about the molecular state of acetic acid in benzene?
Answer» Solution :Here, we are given `m="0.1 mol kg"^(-1),""K_(f)="5.12 K kg mol"^(-1)` `therefore"Theroretically calculated value of "DeltaT_(f)" will be "DeltaT_(f)=K_(f) xx m ="5.12 K kg mol"^(-1)xx"0.1 mol kg"^(-1)=0.512K` Observed (experimental) value of `DeltaT_(f)=0.256K` (Given) Thus, the observed value is HALF of the theoretical value. Since the observed value depends upon the number of particles actually present, this menas that the number of particles actually present in half of the theoretical value. In other words, the molecules of acetic acid are DOUBLY associated in benzene, i.e., they exist as dimers, `(CH_(3)COOH)_(2)`. Alternatively, van't HOFF factor, `i=("Observed colligative property")/("Calculated colligative property")=(0.256)/(0.512)=(1)/(2)` `"Also,i"=("Calculated mol. mass")/("Observed mol. mass")""therefore"Observed mol. mass "=("Calculated mol. mass")/("i")` But calculated molar mass of `CH_(3)COOH="60 g mol"^(-1) ""therefore"Observed molar mass "=(60)/(1//2)="120 g mol"^(-1)` Thus, the observed molar mass is double of the theoretical value. Hence, acetic acid EXISTS as doubly associated, i.e., as `(CH_(3)COOH)_(2)`.