The freezing point of benzene decreases by 0.45^(@)C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (K_(f) for benzene = 5.12 K kg "mol"^(-1))

The freezing point of benzene decreases by 0.45^(@)C when 0.2 g of ace

1.	The freezing point of benzene decreases by 0.45^(@)C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (K_(f) for benzene = 5.12 K kg "mol"^(-1))
Answer» `76.6%` `94.6%` `64.6%` `80.4%` Solution :`DeltaT_(f"(observed)")=0.45^(@)` `DeltaT_(f"(calculated)")=(1000K_(f)w_(2))/(w_(1)xxM_(2))` `=(1000xx5.12xx0.2)/(20xx60)(M_(2)(CH_(3)COOH)=60)` `= 0.853^(@)` `therefore""=(DeltaT_(f"(observed)"))/(DeltaT_(f"(calculated)"))=(0.45)/(0.853)=0.527` `underset(1-alpha)(2CH_(3)COOH)hArr underset((alpha)/(2))((CH_(3)COOH)_(2))` Total moles `=1-alpha+(alpha)/(2)=1-(alpha)/(2)` `i=1-(alpha)/(2)=0.527 or (alpha)/(2)=0.473 or alpha=0.946` `%" association "=94.6%`