Question

When 0.1 mol ch3cooh is present in a litre solvent?

Answer 1

Answer:

CH3COOH⇌CH3COO⊝+H⊕

at t=0 0.1 0 0

at eqm 0.1−C C C

now, i=0.1−C+C+C=0.1+C

ΔTb=iKf×m

⇒0.75=(0.1+C)×5×0.1 [Since M≃m=0.1]

∴ dissociation constant= 5×10−2