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Unformatted text preview: Homework 3 Solutions 1. c (g/dL) time (sec)
1.21261116 0.21261116 2.65763945 2.40970018
1.33185125 0.33185125 2.76542711 2.38808245
1.45432498 0.45432498 2.83953112 2.34088663
1.58367017 0.58367017 2.91835085 2.29872523
1.72635408 0.72635408 3.02647534 2.27504882 3.2
y = 2.4852 + 2.2265x R= 0.99748
y = 2.486 + -0.89665x R= 0.99238 viscosity (dL/g) 3 2.8 (reduced)
(inherent) 2.6 2.4 2.2
0.05 0.1 0.15 0.2 0.25 concentration (g/dL) For both cases, we get =2.49 dL/g. The fact that both lines give the same value is a good check on our procedure. Using the Mark‐Houwink parameters given: K=0.000113 dL/g a=0.73 a KM v Ma v K a 2.49dL / g 0.73
Mv 890,000 0.000113dL / g K 1 1 Note that this is one time that units don't exactly work correctly. Although the answer comes out unitless, it is assumed that the molecular weight units are g/mol. This is because K and a are determined from experiments on samples with known molecular weights, and when K and a are determined this way the units of g/mol are used. 2. a. Mn sample mass # of endgroups If we take 1 L of solution, there are 25.0 g of polymer and 1.00x10‐3 mol of polymer (one endgroup per polymer chain). Therefore, Mn 25.0 g / L
1.00 x10 3 mol / L 2.50 x10 4 g / mol b. This is like part a, except that there are only 0.500x10‐3 mol/L of polymer, because there are two endgroups per chain. Therefore, Mn 25.0 g / L
0.500x10 3 mol / L 5.00x10 4 g / mol 3. Note: Density values in the back of the text are given as ranges. Your answer may be slightly different depending on what you picked within that range. These answers are based on selecting on the density in the middle of the range. a. M = 2X12.011 g/mol + 3x1.008 g/mol + 1x35.453 g/mol = 62.50 g/mol .
. / 43.10 / / /
20.4 Any solvents with a solubility parameter within 2 MPa1/2 are possible. So benzene, chloroform, tetrahydrofuran, chlorobenzene, methylene chloride, dioxane. b. M = 2X12.011 g/mol + 4X1.008 g/mol = 28.05 g/mol . .
∗ / 29.53 / / /
18.4 Possible solvents are carbon tetrachloride, toluene, benzene, chloroform, tetrahydrofuran, chlorobenzene, methylene chloride. c. M = 3X12.011 g/mol + 6X1.008 g/mol + 1X15.999 g/mol = 58.08 g/mol . . / 48.40 . / / /
18.8 Possible solvents are carbon tetrachloride, toluene, benzene, chloroform, tetrahydrofuran, chlorobenzene, methylene chloride, dioxane. 4. Table 3‐3 is based on Hansen three dimensional solubility parameters, which more accurately account for specific interactions than the simpler calculation of problem 1. Since PVC contains permanent dipole secondary bonds, we would expect that the Hansen approach would better account for those specific interactions. 5. A blend would have the lowest entropy of mixing. For mixing of two solvents, each molecule can go anywhere in the lattice. For a polymer with a solvent, each repeat unit of the polymer must be next to the another repeat unit, while the solvent can go anywhere. For two polymers, the repeat units of both polymers must maintain connectivity when placed into the lattice. This restricts the number of possible combinations even more than a polymer plus solvent, so the entropy of mixing is lower. . 6. The morphology is determined by the relative fractions of the number of repeat units in each block: a. Calculating the DP for each block gives 240 for the PS block and 1850 for the PBD block. The PS block is 11% of the total chain length. The morphology is spheres of PS in PBD. b. Calculating the DP for each block gives 960 for the PS block and 7395 for the PBD block. The PS block is 11% of the total chain length. The PS block is 11% of the total chain length. The morphology is spheres of PS in PBD. c. Calculating the DP for each block gives 1920 for the PS block and 1850 for the PBD block. The PS block is 51% of the total chain length. The morphology is lamellae. d. Calculating the DP for each block gives 1920 for the PS block and 3995 for the PMMA block. The PS block is 32% of the total chain length. The morphology is cylinders of PS. 7. Although the morphology is the same in both cases, the blocks are four times as long in part b compared to part a. Since the radii of the spheres depend on the block lengths, the spheres will be bigger for part b compared to part a. ...
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- Fall '10