Given a sample of poly[ethylene-stat-(vinyl acetate)] A.Calculate the mean repeat unit molar mass for a sample of poly[ethylene-stat-(vinyl acetate)] that comprises 12.9 wt% vinyl acetate repeat units.B.Given that its number-average molar mass is 39,870 g/mol, calculate the number-average degree of polymerization of the copolymer.

Answers

Answer 1

Answer:

a) The mean repeat unit molar mass for PEVA is 30.72 g/mol

b) degree of polymerization of the copolymer is 1300

Explanation:

Given that;

the wt% of copolymer consist of 12.9 wt% of vinyl acetate and 87.1 wt% Ethylene.

Basis: 100 g of PEVA consist of 12.9 of vinyl acetate and 87.1g of Ethylene.

now we calculate the mole fraction of vinyl acetate Ethylene in the copolymer;

the molecular weights of vinyl acetate and ethylene are 86.09 g/mol and 28.05 g/mol respectively

so

moles of vinyl acetate = wt. of vinyl acetate / molecular weights of vinyl acetate

moles of vinyl acetate = 12.9 g / 86.09 g/mol

moles of vinyl acetate = 0.1498 mol

moles of Ethylene = wt. of Ethylene / molecular weights of Ethylene

moles of Ethylene = 87.1 g / 28.05  d/mol

moles of Ethylene  = 3.1052 mol

Total moles = 0.1498 mol + 3.1052 mol = 3.255 mol

Next we calculate the mole percent;

mole percent of vinyl acetate [tex]X_{V}[/tex] = moles of vinyl acetate / total  moles

[tex]X_{V}[/tex] = (0.1498 mol / 3.255 mol) × 100

[tex]X_{V}[/tex]  = 4.6%

mole percent of Ethylene [tex]X_{E}[/tex] = moles of Ethylene / total  moles

[tex]X_{E}[/tex]  = (3.1052 mol / 3.255 mol) × 100

[tex]X_{E}[/tex]  = 95.397% ≈ 95.4%

we know that, mean repeat unit molar mass for a sample = ∑[tex]X_{i}[/tex][tex]M_{i}[/tex]

where [tex]X_{i}[/tex] is the fraction ratio and [tex]M_{i}[/tex] is the molecular  weight

so or the PEVA

mean repeat unit molar mass M = ( [tex]X_{V}[/tex][tex]M_{V}[/tex]) + ( [tex]X_{E}[/tex][tex]M_{E}[/tex])

so we substitute

M = ( 4.6% × 86.09) + ( 95.4% × 28.05 )

M = 3.96014 + 26.7597

M = 30.72 g/mol

Therefore, The mean repeat unit molar mass for PEVA is 30.72 g/mol

b)

Degree of polymerization

[tex]DP_{n}[/tex] = [tex]\frac{M_{n} }{M}[/tex]

where [tex]M_{n}[/tex] is the number average molecular weight ( 39,870 g/mol )

so we substitute

[tex]DP_{n}[/tex] = 39,870 g/mol / 30.72 g/mol

[tex]DP_{n}[/tex]  = 1297.85 ≈ 1300   { 3 significance figure }

Therefore, degree of polymerization of the copolymer is 1300


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