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# Chap 16_addendum - Xi = of chains total of chains = I 0,000...

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Solution corrections: 16-10 Analysis of a sanpleof polyacrylonitile (PAI.Q (see Table 16-3) shows that there aresix lengtbs of chains, with the following number of chains of each length. Determine (a) the weight average molecular weight and degree of polymerization and O) the number average molecular weight and degree of polymerization. Number of chains Mean Mol. Weight of chains (g/mol) Mi number fraction xi number average molecular weight xiMi weight weight fraction fi weight average molecular weight fiMi 10,000 3,000 0.1 37 411 3.00E+07 0.044 133 18,000 6,000 0.247 1479 1.08E+08 0.1 59 956 17,000 9,000 0.233 2096 1.53E+08 0.226 2031 15,000 12,000 0.205 2466 1.80E+08 0.265 31 86 9,000 15,000 0.123 1849 1.35E+08 0.1 99 2987 4,000 18,000 0.055 986 7.20E+07 0.1 06 1912 Totals 73,000 63,000 g/mol 1.000 9285 g/mol 6.78E+08 g 1.000 11204 g/mol
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Unformatted text preview: Xi = # of chains / total # of chains = I 0,000 / 73,000 : 0.137 XiMi:0.137 x 3000:411 Weight: (# chains) (Mean Mi per chain) = 10,000 x 3,000 :3.00 E+07 Fi = weight / total weight: 3.008+07 / 6.78E+08 = 0.044 FiMi = (0.044) ( 3000) = 133 Average Mw: sum of fiMi : 11204 DPw: (ln0aglllaol ) / (53 9mol)= 211 Average Mn: sum of xMi :9285 DPn = (9285 g/mol) | (53 glmol) = 175 The molecular weight of the acrylonitrile monomer is M W : 3 C + l N + 3 H : 5 3 g l m o l (a) The weight average molecular weight and degree of polymerization are: MW*: 11204 {mol DP* = ll204l 53:2ll (b) The number average molecular weight and degree of polymerization are: MWn = 9285 g/mol DPn = 9285153: 175...
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