HW 9. Problem a. To Find:


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1 HW 9 Problem To Find: ( The numberverge moleulr weight (b The weightverge moleulr weight ( The degree of polymeriztion for the given polypropylene mteril Moleulr Weight Rnge (g/mol x i w i 8,000 16, ,000 24, ,000 32, ,000 40, ,000 48, ,000 56, The given dt is urte; the mteril (polypropylene is pure. ( Numberverge moleulr weight: Moleulr wt. Rnge Men M i x i x i M i 8,00016,000 12, ,00024,000 20, ,00032,000 28, ,00040,000 36, ,080 40,00048,000 44, ,00056,000 52, M n = x i M i = 33,040 g/mol
2 (b Weightverge moleulr weight: Moleulr wt. Rnge Men M i w i w i M i 8,00016,000 12, ,00024,000 20, ,00032,000 28, ,00040,000 36, ,800 40,00048,000 44, ,880 48,00056,000 52, ( Degree of polymeriztion: M w = w i M i = 36,240 g/mol For polypropylene, the repet unit moleulr weight, m = 3(A C + 6(A H = (3(12.01 g/mol + (6(1.008 g/mol = g/mol DP = M n m 33,040 g/mol = g/mol = 785 ( g/mol (b g/mol (
3 Problem To Find: ( Wt% of Chlorine to be dded (b How this hlorinted polyethylene differs from PVC Highdensity polyethylene is hlorinted suh tht Cl toms reple 5% of the originl H toms. (i (ii (iii (iv All the Chlorine dded is utilized in repling the existing H toms no Cl tom stys unutilized. Extly 5 % of the originl H toms re repled by Cl toms. The originl mteril is pure polyethylene. No other tom speies (whih ould potentilly reple H toms is dded. ( Consider 50 rbon toms in the high density polyethylene. These orrespond to 100 possible sidebonding sites. Initilly, ll 100 sites re oupied by H. Posthlorintion, 95 of these sites re oupied by hydrogen nd 5 sites re oupied by Cl. Mss of 50 rbon toms,m C = 50(A C = (50(12.01 g/mol = g Mss of 95 H toms, m H = 95(A H = (95(1.008 g/mol = g Mss of 95 Cl toms, m Cl = 5(A Cl = (5(35.45 g/mol = g Using modified form of Eqution 4.3, onentrtion of hlorine, C Cl is: C Cl = m Cl x 100 = m C + m H + m Cl g g g g 100 = 20.3 wt% 20.3 wt% (b Compred to hlorinted polyethylene, in poly(vinyl hloride: (1 25% of the sidebonding sites re substituted with Cl (2 Substitution is probbly less rndom
4 Problem To Find: The numberverge moleulr weight of rndom nitrile rubber [poly(rylonitrilebutdiene opolymer]. 1. Frtion of butdiene repet units, f Bu = 0.30 (=> frtion of rylonitrile repet units, f A = Degree of polymeriztion = 2000 Given dt is urte; the mteril is pure. The two repet units in this opolymer re rylonitrile nd butdiene. From Tble 14.5, the rylonitrile repet unit ontins 3 Crbon toms, 1 Nitrogen tom nd 3 Hydrogen toms. Thus, the moleulr weight of the rylonitrile repet unit is: m A = 3(A C + (A N + 3(A H = (3(12.01 g/mol g/mol + (3(1.008 g/mol = g/mol The butdiene repet unit ontins 4 Crbon toms nd 6 Hydrogen toms. Thus, the moleulr weight of the butdiene repet unit is: m Bu = 4(A C + 6(A H = (4(12.01 g/mol + (6(1.008 g/mol = g/mol From Eqution 14.7, the verge repet unit moleulr weight is: m = f A m A + f Bu m Bu = (0.70(53.06 g/mol + (0.30(54.09 g/mol = g/mol From Eqution 14.6, the numberverge moleulr weight is: M n = m (DP = (53.37 g/mol(2000 = 106,740 g/mol g/mol
5 Problem To Find: ( Rtio of butdiene to styrene repet units in the given opolymer (b Type of opolymer 1. Numberverge moleulr weight of opolymer = g/mol 2. Degree of polymeriztion = 4425 (i (ii Given dt is urte. The mteril is pure. ( From Eqution 14.6, the verge repet unit moleulr weight of the opolymer, lulted s: m = M n DP 350,000 g/mol = 4425 = g/mol m, is From Tble 14.5, the butdiene repet unit ontins 4 Crbon toms nd 6 Hydrogen toms. Thus, the moleulr weight of the butdiene repet unit is : m b = 4(A C + 6(A H = 4(12.01 g/mol + 6(1.008 g/mol = g/mol The styrene repet unit ontins 8 Crbon toms nd 8 Hydrogen toms. Thus, the moleulr weight of the styrene repet unit is: m s = 8(A C + 8(A H = 8(12.01 g/mol + 8(1.008 g/mol = g/mol Let f b be the hin frtion of butdiene repet units. Sine there re only two repet unit types in the opolymer, the hin frtion of styrene repet units f s is 1 f b. Eqution 14.7 my be written s: m = f b m b + f s m s = f b m b + (1 f b m s f b = m m s m b m s f s = 1 f b = = 0.50 = g/mol g/mol g/mol g/mol = 0.50
6 Hene, the rtio of butdiene to styrene repet units = f b = 0.50 f s 0.50 = 1.0 (or 1:1 (b The rtio of the the two units in the opolymer is 1:1 in se of lternting opolymers. There is no fixed rtio presribed for rndom, grft nd blok opolymers (1:1 ould be possible rtio for eh of these types! Therefore, with the rtio of butdiene to styrene repet units obtined in prt (, this opolymer be of lternting, rndom, grft nd blok types. ( 1 (or 1:1 (b Alternting, rndom, grft nd blok types
7 Problem To Find: ( Densities of totlly rystlline nd totlly morphous polytetrfluoroethylene (PTFE. (b % Crystllinity of speimen hving density of 2.26 g/m 3. ρ (g/m 3 rystllinity (% Given dt is urte; mteril is pure. s ( From Eqution 14.8: %Crystllinity = * 100 s ( ( (1 Where is the density of totlly rystlline PTFE; is the density of totlly morphous PTFE nd s is the density of the given speimen. We n obtin two equtions from the two sets of dt given nd solve for nd. (A For s = nd % rystllinity = = * ( => 51.3*2.144/100 = ( (2 (B For s = nd % rystllinity = = * ( * / 100 = ( (3
8 Dividing (2 by (3 : 51.3*2.144 / (74.2*2.215 = ( ( = ( ( = g/ m 3 => = => = Substituting the vlue of in (2 : 51.3*2.144/100 = ( = ( = g/ m 3 => = => = 2.2 (b Substituting the vlues of nd in (1 for s = 2.26 g/m 3 : % Crystllinity = (2.301 g/m3 ( g/m g/m 3 (2.260 g/m 3 ( g/m g/m 3 100= 87.9 ( = g/ m 3 ; = g/ m 3 (b 87.9 % Crystllinity
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