WB=2g,
Kf=4.9K Kgmol−1
WA=25g,
ΔTf=1.62K
now
If x is the degree of association, (1−x) mole of benzoic acid left undissociated & corresponding x/2 as associated moles of C6H5COOH at equilibrium.

i = Normal molecular mass / Abnormal molecular mass
1-x/2 = 122/241.98
x=0.992 = 99.2%
