Question

The freezing point of a solution containg 0.2 gm acetic acid in 20g of benzene

Answer 1

Given :

w=0.2g

W=20g

ΔT=0.45

Applying,ΔT=1000×Kf×wm×W

Or,0.45=1000×5.12×0.220×m

∴m(observed)=113.78(acetic acid)

2CH3COOH⇋(CH3COOH)2

Before dissociation 10

After dissociation 1−αα2

α-Degree of dissociation

Molecular weight of acetic acid =60

i=normal molecular massobserved molecular mass

∴mnormalmobserved=1−α+α2

60113.78=1−α+α2

∴α=0.945