QJEEAdv06Chde33: JEE Advanced 2006 Chemistry - Class 12 JEEAdv06Chde33 Solution: 75.2 g of C 6 H 5 OH (phenol) is dissolved in a solvent of K f…

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

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

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

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

$$\therefore$$ $$\mathrm{\Delta }$$T_f = 14 $$\times$$ 0.77 = 10.78 K

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

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

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

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

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

In case of dimerisation, n = 2

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

Hence, the percentage of phenol that dimerises = 70%