Question

Acetic acid dimerises in benzene solution. The van't Hoff factor for dimerisatiion of acetic acid is 0.6. The % of dimerisation of acetic acid is

Choose the correct answer:

 * 20%
  *60%
  *80%
  *100%

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Answer 1

Answer:

The answer is 80%.

Explanation:

Acetic acid associate in benzene to form dimeric molecules.

The equilibrium in the solution is.

2CH3COOH↔↔(CH3COOH)2

Let the degree of dimerization of the solute is x, then we have (1-x) mol of acetic acid left unassociated in the solution, and x/2 moles of acetic acid at equilibrium.

Therefore the number of particles at equilibrium  

 = 1-x +x / 2

 = 1-x2x2

The total number of particles at equilibrium is equal to the Vant Hoff factor. But Vant Hoff factor =0.6

Thus,

1-x / 2=0.6

1-0.6 = x / 2

0.4  = x / 2

x = 0.8

Degree of dimerisation of acetic acid =80 %

Answer 2

Answer:

Answer is 80%.

During the reaction of dimerization in benzene solution the following reaction will be 2CH3COOH → (CH3COOH)2  

And the relation between Van’t Hoff factor which is represented by ‘i’ and degree of dissociation which is represented by beta (β). It can be written as a following equation which is—

i = 1+ (1/n – 1) β and in question i is given which is 0.6 and n is 2. Put this value in the following equation

0.6 = 1 + (1/2-1) β

1/2β = 0.4  

β = 0.8 Now we can calculate the percentage of dimerization which will be β X 100 = 80%