We have a polyethylene (C_2H_4) sample containing 4,000 chains with molecular we
ID: 491715 • Letter: W
Question
We have a polyethylene (C_2H_4) sample containing 4,000 chains with molecular weights between 0 g/mol and 5,000 g/mol, 8,000 chains with molecular weights between 5,000 g/mol and 10,000 g/mol, 7,000 chains with molecular weights between 10,000 g/mol and 15,000 g/mol, and 2,000 chains with molecular weights between 15,000 g/mol and 20,000 g/mol. The atomic mass of hydrogen is 1 g/mol and atomic mass of carbon is 12.01 g/mol. Determine the: (a) Number average molecular weight (b) Weight average molecular weights (c) Number average degree of polymerization (d) Weight average degree of polymerization (e) Polydispersity index.Explanation / Answer
We have a polyethylene sample containing 4000 chains with molecular weights between 0 and 5000 g/mol, 8000 chains with molecular weights between 5000 and 10,000 g/mol, 7000 chains with molecular weights between 10,000 and 15,000 g/mol, and 2000 chains with molecular weights between 15,000 and 20,000 g/mol.
Determine both the number and weight average molecular weights.
SOLUTION
First we need to determine the number fraction xi and weight fraction fi for each of the four ranges.
Number of chain
Mean M Per chain
X1
X1Mi
Weight
fi
fiMi
4000
2500
0.191
477.5
10x10^6
0.0519
129.75
8000
7500
0.381
2857.5
60x10^6
0.3118
2338.50
7000
12500
0.333
4162.5
87.5 x10^6
0.4545
5681.25
2000
17500
0.095
1662.5
35x10^6
0.1818
3181.5
Sum=21000
Sum=1
=9160
=192.5x10^6
=1
11.331
Mn ==9160 g/mol ---------------answer
Mw ==11331 g/mol ---------------answer
Degree of polymerization -
The average molecular weight of the polymer divided by the molecular weight of the monomer
Number of chain
Mean M Per chain
X1
X1Mi
Weight
fi
fiMi
4000
2500
0.191
477.5
10x10^6
0.0519
129.75
8000
7500
0.381
2857.5
60x10^6
0.3118
2338.50
7000
12500
0.333
4162.5
87.5 x10^6
0.4545
5681.25
2000
17500
0.095
1662.5
35x10^6
0.1818
3181.5
Sum=21000
Sum=1
=9160
=192.5x10^6
=1
11.331
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