Question

equimolar amount of monomer A and B are copolymerized . What kind of copolymerization take place if r1=5 and r2= 0.2 ? show that the copolymer initially formed contain 16.7 mol % of monomer B .

Answer 1

Here you can observe that r₁r₂ = 1
5 * 0.2 = 1 ; r₂ = 1 / r₁
Ideal Co-Polymerization (Ionic Polymerization) takes place which results in a random Copolymer.
The copolymer equation gets reduced to,
F[M₁] / F[M₂] = r₁[M₁] / [M₂]
since it is equimolar [M₁] / [M₂] = 1
F[M₁] / F[M₂] = 5
ratio of A to B i.e) A : B = 5 : 1
The mol fraction should add up to give 1 so if B = 0.167 then A = 5*0.167 = 0.835
mol fraction of A + mol fraction of B = 0.835 + 0.167 = 1.002 ( ~ 1 approx)
Hence we can say that the copolymer initially formed contain 16.7 mol % of monomer B .