Question

What is the molecular weight of the polymer prepared by the self
condensation of HO‐C6H5‐COOH, if 99.5% of the functional groups react

Answer 1

Let there be N monomers

one of out N monomers will have free COOH , 99.5 % reacted which means 0.5 % unreacted .

So the free COOH of one monomer out of N monomers is 0.5 %

hence ( 1/N) = 0.5 %

1/N = ( 0.5/100)

N = ( 100 / 0.5) = 200

Thus we had 200 monomers

when two monomers combine we get dimer with elimination of H2O of molar mass 18

when 3 monomers combine 2 H2O eliminated

hence when 200 monomers combine 199 H2O eliminated which means 199 x 18 = 3582 g/mol

Total molar mass of polymer = ( 200 x monomer molar mass) - ( molar mass of 199 H2O eliminated)

= ( 200 x 138 g/mol) - 3582g/mol

= 24018 g/mol