SOLUTIONS OF EXERCISES IN LECTURE 8 Exercise #1 Polystyrene: M 0 =104 g.mol -1 ; -1 g.mol 89440 = n M ; PDI = 1.5 860 = n x and -1 g.mol 134160 = w M Exercise #2 Poly(ethylene-stat-vinyl acetate); vinyl acetate: va; ethylene: et M et =28 g.mol -1 ; M va =86 g.mol -1 ; -1 g.mol 39870 = n M Mole fraction of ethylene, n et = [(87.1/100)/28]/((0.871/28)+(0.129/86)) = 0.954 Mole fraction of vinyl acetate, n va = 0.046 1 0 . 67 . 30 28 954 . 0 86 046 . 0 − = + = mol g x x M cop 1300 = n x Exercise #3 MW (kg.mol -1 ) n i ) (kg.mol -1 n M w i ) (kg.mol -1 w M PDI w x 9 0.05 0.45 0.008392 0.075532 1.107 1349 32 0.1 3.2 0.059679 1.909735 41 0.21 8.61 0.160574 6.583551 56 0.31 17.36 0.32376 18.13055 70 0.23 16.1 0.300261 21.01828 79 0.10 7.9 0.14733 11.63931 53.62 59.37 Exercise #4 MW (g.mol -1 ) W(g) w i ) (g.mol -1 w M # of moles n i (g.mol - n M PDI 70000 1 0.0833 5,833.33 1.42 x 10 -5 0.01958 1,370.75 1.49 43000 2 0.1666 7,166.66 4.65 x 10 -5 0.06376 2,741.50 19000 6 0.5 9,500 3.15 x 10 -4 0.43287 8,224.50 8500 3 0.25 2,125 3.52 x 10 -4 0.48379 4,112.25 12 2,4625 16,448.99 O O CH 3 n m MATS3004/6109 6/16/21 SOLUTIONS OF EXERCISES IN LECTURE 8 Exercise #5 Fraction Weight (g) 𝑴𝑴 𝒏𝒏 (g.mol -1 )* Weight fraction, (w i ) 𝑴𝑴 � 𝒏𝒏 = 𝟏𝟏 ∑� 𝒘𝒘 𝒊𝒊 𝑴𝑴 𝒊𝒊 � (g.mol -1 ) 𝑴𝑴 𝒘𝒘 (g.mol -1 )* 1 1.5 2,000 0.032 63.83 2 5.5 50,000 0.117 5851.06 3 22 100,000 0.468 46808.51 4 12 200,000 0.255 51063.83 5 4.5 500,000 0.096 47872.34 6 1.5 1,000,000 0.032 31914.89 47 40851.80 183574.5 * the given assumes each fraction is monodisperse, hence we will refer to 𝑀𝑀 � 𝑛𝑛 as M i