Neco 2024 Chemistry Question 30

A step‑growth polymer is formed from equimolar amounts of monomer A (\(M_{\text{w}} = 100\ \text{g mol}^{-1}\)) and monomer B (\(M_{\text{w}} = 150\ \text{g mol}^{-1}\)). The extent of reaction (conversion) is \(p = 0.95\).\n\n(a) Calculate the number‑average degree of polymerisation \(DP_{\text{n}}\) and the number‑average molecular weight \(M_{\text{n}}\).\n(b) The weight‑average degree of polymerisation is given by \(DP_{\text{w}} = \frac{1+p}{1-p}\). Compute \(DP_{\text{w}}\) and the weight‑average molecular weight \(M_{\text{w}}\). Determine the polydispersity index \(PDI = DP_{\text{w}}/DP_{\text{n}}\).\n(c) Tensile strength \(\sigma\) of the polymer is proportional to \(1/\sqrt{DP_{\text{n}}}\). If a polymer with \(DP_{\text{n}} = 100\) has a tensile strength of \(30\ \text{MPa}\), estimate the tensile strength of the polymer obtained above.