Question

A solution of phenol was obtained by dissolving 2 × 10^-2 kg of phenol in 1 kg of benzene. Experimentally, it was found to be 73% associated. Calculate the depression in the freezing point recorded.

Answer 1

\[\ce{2C6H5OH -> (C6H5OH)2}\]

Initial concentration	C	0
Final concentration	C (1 − α)	Cα/n

where α is degree of association.

Experimentally, phenol is 73% associated.

Hence α = 0.73

Relation between i (vant hoff factor) and α is given as: α = `(1 − i)/(1 − n)`, where n for phenol = `1/2` as phenol acts as dimer, association is taking place

Substituting the values:

0.73 = `(1 − i)/(0.5)`

i = 1 − (0.73 × 0.5)

i = 1 − 0.365

i = 0.635

Depression in freezing point can be calculated as:

K_f = 5.12 K Kg/mol,

W_b = 2 × 10^-2 kg = 20 g,

W_a = 1 kg = 1000

M_b = 94

ΔT_f = `(k_f xx i xx W_b)/M_b`

= `(5.12 xx 0.635 xx 20)/94`

= `65.024/94`

= 0.691 K