[sorina.l]

Hello,

I am trying to solve a problem and I am not sure how, but I am close. I just need a hint. It sounds like that: Let's consider the living anionic polymerization of styrene. Choose the initial concentration of monomer and initiator so that a polymer with a number average molecular weight of 10400 be obtained at full monomer conversion. The contribution of the initiator to the molecular weight of the polymer is neglected.
The following formula must be used:
DPn=((n x M0)/I0) x Styrene molecular weight   where DPn-the nr. average degree of polymerization;
                                                                                n-the conversion;
                                                                                M0- the initial concentration of monomer;
                                                                                I0- the initial concentration of initiator;
If I0 is neglected we must consider it 1 mol/L in the equation?
If we consider that, then 10400=M0 x Styrene molecular weight (104) --> M0=100 mol/L and I0=1 mol/L
Do you think it is a good solving? Thank you!

[mjc123]

You are confusing molecular weight with degree of polymerisation.
DPn = (n x M0)/I0
Mn = DPn x Styrene molecular weight
You are correct that for Mn = 10400 we want M0 = 100 x I0, but a concentration of 100 mol/L is totally unrealistic. A litre of pure styrene is about 9 moles.

[sorina.l]

Thank you @mjc123. I understand now