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Chapter 14: Problem 5

The following table lists molecular weight data for a polypropylene material. Compute (a) the number-average molecular weight, (b) the weight-average molecular weight, and (c) the degree of polymerization. \begin{tabular}{rcc} \hline Molecular Weight Range \((\mathrm{g} /\) mol \()\) & \(\boldsymbol{x}_{\boldsymbol{i}}\) & \(\boldsymbol{w}_{\boldsymbol{i}}\) \\ \hline \(8,000-16,000\) & \(0.05\) & \(0.02\) \\ \(16,000-24,000\) & \(0.16\) & \(0.10\) \\ \(24,000-32,000\) & \(0.24\) & \(0.20\) \\ \(32,000-40,000\) & \(0.28\) & \(0.30\) \\ \(40,000-48,000\) & \(0.20\) & \(0.27\) \\ \(48,000-56,000\) & \(0.07\) & \(0.11\) \\ \hline \end{tabular}

Short Answer

Expert verified
Question: Calculate the number-average molecular weight, weight-average molecular weight, and the degree of polymerization for a given polypropylene material based on the provided molecular weight data. Answer: The number-average molecular weight (M_n) is 31,920 g/mol, the weight-average molecular weight (M_w) is 32,736 g/mol, and the degree of polymerization (X_n) is 760.

Step by step solution

01

Calculate Average Molecular Weight

To calculate the average molecular weight for each range, it's necessary to take the midpoint value for the range. For the given ranges in the table: Range 1: (8000+16000)/2 = 12000 g/mol Range 2: (16000+24000)/2 = 20000 g/mol Range 3: (24000+32000)/2 = 28000 g/mol Range 4: (32000+40000)/2 = 36000 g/mol Range 5: (40000+48000)/2 = 44000 g/mol Range 6: (48000+56000)/2 = 52000 g/mol
02

Calculate Number-Average Molecular Weight

To calculate the number-average molecular weight (M_n), use the formula: \(M_{n} = \sum_{i=1}^{n} x_{i}M_{i}\) where \(x_{i}\) is the mole fraction and \(M_{i}\) is the average molecular weight for each range. Using the given values: \(M_n = 0.05(12000) + 0.16(20000) + 0.24(28000) + 0.28(36000) + 0.20(44000) + 0.07(52000)\) \(M_n = 31920 \thinspace g/mol\)
03

Calculate Weight-Average Molecular Weight

To calculate the weight-average molecular weight (M_w), use the formula: \(M_{w} = \sum_{i=1}^{n} w_{i}M_{i}\) where \(w_{i}\) is the weight fraction and \(M_{i}\) is the average molecular weight for each range. Using the given values: \(M_w = 0.02(12000) + 0.1(20000) + 0.2(28000) + 0.3(36000) + 0.27(44000) + 0.11(52000)\) \(M_w = 32736 \thinspace g/mol\)
04

Calculate Degree of Polymerization

The degree of polymerization (X_n) is given by the formula: \(X_n = \dfrac{M_n}{M_{\text{monomer}}}\) Considering the molar mass of the propylene monomer is 42.0 g/mol. Calculating the degree of polymerization: \(X_n = \dfrac{31920}{42.0}\) \(X_n = 760\)

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Most popular questions from this chapter

The density and associated percent crystallinity for two polytetrafluoroethylene materials are as follows: \begin{tabular}{cc} \hline\(\rho\left(\mathrm{g} / \mathrm{cm}^{3}\right)\) & crystallinity \((\%)\) \\\ \hline \(2.144\) & \(51.3\) \\ \(2.215\) & \(74.2\) \\ \hline \end{tabular} (a) Compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene. (b) Determine the percent crystallinity of a specimen having a density of \(2.26 \mathrm{~g} / \mathrm{cm}^{3}\).

For each of the following pairs of polymers, do the following: (1) state whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Linear and syndiotactic poly(vinyl chloride); linear and isotactic polystyrene (b) Network phenol-formaldehyde; linear and heavily crosslinked \(c i s\)-isoprene (c) Linear polyethylene; lightly branched isotactic polypropylene (d) Alternating poly(styrene-ethylene) copolymer; random poly(vinyl chloridetetrafluoroethylene) copolymer

Using the definitions for total chain molecule length \(L\) (Equation 14.11) and average chain end-to-end distance \(r\) (Equation 14.12), for a linear polyethylene determine the following: (a) the number-average molecular weight for \(L=2500 \mathrm{~nm}\) (b) the number-average molecular weight for \(r=20 \mathrm{~nm}\)

For a linear freely rotating polymer molecule, the total extended chain length \(L\) depends on the bond length between chain atoms \(d\), the total number of bonds in the molecule \(N\), and the angle between adjacent backbone chain atoms \(\theta\), as follows: $$ L=N d \sin \left(\frac{\theta}{2}\right) $$ Furthermore, the average end-to-end distance \(r\) for a randomly winding polymer molecule in Figure \(14.6\) is equal to $$ r=d \sqrt{N} $$ A linear polytetrafluoroethylene has a numberaverage molecular weight of \(500,000 \mathrm{~g} / \mathrm{mol}\); compute average values of \(L\) and \(r\) for this material.

Crosslinked copolymers consisting of \(60 \mathrm{wt} \%\) ethylene and \(40 \mathrm{wt} \%\) propylene may have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both repeat unit types.
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