Gce Model Questions Vol1 2017 Chemistry Question 31

A step‑growth condensation polymerisation is carried out between adipic acid \(\mathrm{C_6H_{10}O_4}\) (M = 146.14 g·mol⁻¹) and hexamethylenediamine \(\mathrm{C_6H_{16}N_2}\) (M = 116.21 g·mol⁻¹) to give nylon‑6,6 and water. Initially 0.500 mol of each monomer are placed in the reactor. The reaction proceeds until 92 % of each monomer has reacted, but 5 % of the polymer chains formed undergo intramolecular cyclisation, removing them from the linear polymer population. \(\text{(a)}\) Calculate the number‑average degree of polymerisation \(DP_n\) of the linear polymer present. \(\text{(b)}\) Using the repeat‑unit molar mass \(M_{RU}=226.34\ \text{g·mol}^{-1}\) (the sum of the two monomers minus the mass of one water molecule), determine the number‑average molecular weight \(M_n\) of the linear polymer.