Question

72.5 g of phenol is dissolved in 1 kg of solvent (kf=14) depression in freezing point is 7 k find the % of phenol that dimersises

Answer 1

Mb =  

Taking the solvent as 1 kg

Mb =  

= 150.4 g per mol

phenol (molar mass) — 94 g/mol

i =  

i = 94 / 150.4 = 0.625

 

We have, 2C6H5OH ↔ (C6H5OH)2

1 0

1- α α/2

Total = 1- α + α/2

i = 1- α/2

0.625 = 1- α/2

α/2 = 0.375

α = 0.375/2

α = 0.75

% of phenol that dimerizes = 75%

Answer 2

Hello Student,

Please find the answer to your question

2C6H5OH ⇌ (C6H5OH)2

Initial no. of moles 1 0

No. of moles at equilibrium 1 - ∝ ∝/2

Total number of moles at equilibrium = 1 - ∝ + ∝/2 = 1 - ∝/2

∆Tf = iKf* (molality)

⇒ 7 = 14 * 75.2/94 * (1 - ∝/2) [weight of phenol = 75.2g mol. wt of phenol = 94]

∴ ∝ = 0.75

So the percentage of phenol that dimerises = 75%.

ALTERNATIVESOLUTION :

Phenol dimerises in the solvent (organic) as :

2C6H5OH ⇌ (C6H5OH)2

1 mol 0 mol (Initial)

1 - ∝ mol ∝/2 mol (Equilibrium)

Van’t Hoff factor, i = 1 - ∝ + ∝/2 = 1 - ∝/2

Where ∝ is the degree of dimerization.

∆Tf(obs) = Kf * molality * i

Or 7 = 14 * m * (1 - ∝/2)

NOTE : In the given problem, amount of the solvent used to dissolved 75.2 g of phenol is not mentioned. Hence molality m, can not be calculated and so also the value of ∝.

If amount of solution is presumed to be 1 kg (= 100g) then,

M = 75.2/94 = 0.8 (mol. mass of phenol = 9.4)

And 7 = 14 * 75.2/94 * (1 - ∝/2)

Or ∝ = 0.75 = 75%