No. of moles of dissolved phenol in solution = $${{72.5} \over {94}}$$ = 0.77 mol

[$$\because$$ molar mass of phenol = 94 g mol^-1
]

Molality (m) of the solution = $${{0.77} \over {1000}} \times 1000 = 0.77$$ mol kg^$$-$$1 [$$\because$$ weight of solvent = 1000 g]

We know, $$\Delta$$T_f = k_f $$\times$$ m [Given : k_f = 14]

$$\therefore$$ Delta$$T_f = 14 $$\times$$ 0.77 = 10.78 K

Hence, calculated value of $$\Delta$$T_f = 10.78 K

But, experimental value of $$\Delta$$T_f = 7K

Therefore, van't Hoff factor, $$i = {7 \over {10.78}} = 0.65$$

If a solute undergoes association in solution, its degree of association ($$\alpha$$) is given by, $$\alpha = {{i - 1} \over {{1 \over n} - 1}}$$

where n = no. of molecules that associate to form a single associated molecule.

In case of dimerisation, n = 2

$$\therefore$$ $$\alpha = {{0.65 - 1} \over {{1 \over 2} - 1}} = 0.7$$

Hence, the percentage of phenol that dimerises = 70%