A polymer has the following composition: 100 molecules of molecular weight 1500, 200 molecules of molecular weight 2500, and 500 molecules of molecular weight 5000. Calculate the number and weight average molecular weight and the polydispersity index.
Question
A polymer has the following composition: 100 molecules of molecular weight 1500, 200 molecules of molecular weight 2500, and 500 molecules of molecular weight 5000. Calculate the number and weight average molecular weight and the polydispersity index.
Solution
To calculate the number average molecular weight (Mn), weight average molecular weight (Mw), and the polydispersity index (PDI), we need to follow these steps:
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Calculate the total number of molecules: Total number of molecules = 100 + 200 + 500 = 800 molecules
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Calculate the number average molecular weight (Mn): Mn = (Sum of (Ni * Mi)) / Total number of molecules Where Ni is the number of molecules of each type and Mi is the molecular weight of each type. Mn = [(100 * 1500) + (200 * 2500) + (500 * 5000)] / 800 Mn = 4125 g/mol
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Calculate the weight average molecular weight (Mw): Mw = Sum of (Wi * Mi) Where Wi is the weight fraction of each type of molecule and Mi is the molecular weight of each type. Wi = Ni * Mi / Sum of (Ni * Mi) Mw = [(100 * 1500 * 1500) + (200 * 2500 * 2500) + (500 * 5000 * 5000)] / [(100 * 1500) + (200 * 2500) + (500 * 5000)] Mw = 4875 g/mol
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Calculate the polydispersity index (PDI): PDI = Mw / Mn PDI = 4875 / 4125 PDI = 1.18
So, the number average molecular weight is 4125 g/mol, the weight average molecular weight is 4875 g/mol, and the polydispersity index is 1.18.
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Instructions: Use the atomic weights for the elements in your Equation Booklet and the following figure to compute problem 1 seen below.--Given Values--- = Data for Problem 1Molecular Weight Data Table for = Polystyrene (PS)- = Molecular Wt(g/mol)|----xi--- |-- wi- = 8,000 - 16,000 -------|-- 0.03 --|--0.04- = 16,000 - 24,000 ------|-- 0.16 --|--0.12- = 24,000 - 32,000 ------|-- 0.25 --|--0.20- = 32,000 - 40,000 ------|-- 0.29 --|--0.28- = 40,000 - 48,000 ------|-- 0.19 --|--0.29- = 48,000 - 56,000------|-- 0.08 --|--0.13- = Data for Problem 2Required Tensile Strength (MPa) = 154Tensile Strength (MPa) for PE 1 = 77Tensile Strength (MPa) for PE 2 = 160Number-Average Molecular Weight (g/mol) for PE 1 = 17,886Number-Average Molecular Weight (g/mol) for PE 2 = 26,247Problem 1 Part A :For the polymer listed, given the mass range in (g/mol) and Ni (number fraction) and wi(weight fraction). Compute the number - average molecular weight for the polymer. Your Answer =Problem 1 Part B:Calculate the weight - average molecular weight for the polymer Your Answer =Problem 1 Part C:The degree of polymerization of for the polymer listed. Your Answer =Problem 2 Part A:The tensile strength and number-average molecular weight for two polyethylene materials given above. Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa Your Answer =Problem 2 Part B:Using the data given in the problem, find A in Mpa-g/mol. Your Answer =Problem 2 Part C:What is the number-average molecular weight that is required to give a tensile strength required above? Your Answer =
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