Given, In benzene,  2C6H5COOH⟶(C6H5COOH)2 Initial moles   1      0 Final moles   1−α    α/2 Total number of moles of equilibrium =1−α+α/2 =1−α/2
∴i=‌
‌ Total moles at equilibrium ‌
‌ Initial moles ‌
=‌
1−α/2
1
We know that, ∆Tf=iKfm
2⋅2=(1−‌
α
2
)×5.0k‌‌kgmol−1×‌
2.44
122
×‌
1000
25
1−‌
α
2
=‌
2.2×122×25
244×1000×5
⇒1−‌
α
2
=0.55 α=0.90 Percentage degree of association =90%