Question

the van' t Hoff factor of a species undergoing 50% association is 0.67. Then the no. of monomer units in the associated compound is​

Answer 1

Here is your answer

Answer 2

Given : The Van't Hoff factor of a species undergoing 50% association is 0.67.

To find : The no of monomer units in the associated compound.

solution : n(monomer) ⇔polymer

1 - α α/n

⇒i = 1 - α + α/n

here Van't Hoff factor, i = 0.67 , α = 50% = 0.5

so, 0.67 = 1 - 0.5 + 0.5/n

⇒0.67 = 0.5 + 0.5/n

⇒0.17 = 0.5/n

⇒n = 0.5/0.17 ≈ 3

Therefore no of monomer units in the associated compound is 3.