Question

8.a) A random poly(styrene-butadiene) copolymer has a weight-average molecular weight of 467,500 g/mol and a weight-average degree of polymerization of 5500. Calculate the fraction of styrene (C:Hg) and butadiene (C4H6) mers in this copolymer. Atomic weights: C-12, H-1 b) What weight % sulphur must be added to vulcanize polyisoprene (mer weight = 68) by crosslinking one half the possible sites. In this elastomer, each pair of mers has two possible crosslink sites. Assume three sulphur atoms are involved in each crosslink. Atomic weight: S-32

Answer 1

Solution : page of 8 a) The styrene repeat unit has 8 carbon atoms and 8 hydrogen atoms while butadiene repeat unit has a carbon atoms and 6 hydrogen atons Unit molecular weight of styrene mg = n(ActrhAH nc=8; Az=12.01. glud, naa8; A4=1.008 almol ms: 8 x 12001 +881.068 = 104.14 g/mol Unit molecular weight of butadiene → monete AnhAH Me= 4; A2=1,01 g/mol; n = 6; A7 1.088 g/mol me a letto 4x12.01 + 6X1.008 = 54-099/mol Average repeat unit molecular Copolymer ñ = foms +fomb fg mg =) ñ = 104.14 fs +(r-fs) 5409 weight of the + (1-6s) me

page of Degree of polymerization bP- Mo m Mn = 467500 g/mol; M DP= 5500 => 51500 467500 . 104-14f5+(1-4) 54.09 > 104.14 fs +54.09- 54.094s = 467500 SSOO ) 50.05fs = 85- 54.09 ㅕ fs = 30.91 50.05 fs =0.62 fraction of styren rene repeat units is 0.62 Estfo=1 =) 0:6 2 + fo=1 = fo = 1-0.62 - fo= 0.38 os fraction of butadtene repeat inits is 0.38

b ) pogeo of o c H H isopene isoprene molecular weight = 689 /mol CH₃ C c = c 3/4 poly 3/4 c H Jo H isoprene 100). 3 X x 32g 50% Quo 37 32 x So lo - Amount required = 48 H8 g. cs Wt V. 48 required X100 29.3wter. (68+3x32 és 29-3wt% sulphus must be addede