Question

3. What is the no. average degree of polymerization (DP) of each of the following polymers with M, 254,000? но th-chatti itomaan -O-CH2-CH2-0-C- ooo į

Answer 1

Number-average degree of polymerization is given by,

$\huge DP= \frac{M_{n}}{M_{0}}$

Mn is the number-average molecular weight and M0 is the molecular weight of the monomer unit .

Here, Mn = 254000 (given value)

Therefore,

Molecular weight = A+Nchitat: 1N + 6C +11H +1 O = 113 O= B to-CH2-CH2--0-0- Molecular weight = 10 C+8 H+40 = 192 O Molecular weight = 20 +1 C +2 H = 85 sitout. Molecular weight = 201+2C+2+ = 45 Dito atool Molecular weight = 16C+30+ 14H = 254 0- Molecular weight = 16 C+30 + 14H = 254

DP for A = (254000/113) = 2247.79

DP for B = (254000/192) = 1322.92

DP for C = (254000/85) = 2988.24

DP for D   = (254000/254) = 1000