Question

Compute the number-average molecular weight for a polypropylene for which the degree of polymerization is 25,000.

Answer 1

Given:

Propylene is
`C_(3)H_(6)`

Degree of polymerization,
`D_(p)` = 25000

Solution:

no. of carbon atoms,
`n_(1)` = 3

no. of hydrogen atoms,
`n_(2)` = 6

atomic weight of carbon,
`A_(c)` =12.01 g/mol

atomic weight of hydrogen,
`A_(h)` = 1.008 g/mol

molecular mass of propylene, m =
`n_(1)A_(c)+n_(2)A_(h)`

m =
`3* 12.01+6* 1.008 = 42.08` g/mol

Now,

`D_(p)` =
`\frac{\bar{M_(n)}}{m}`

`{\bar{M_(n)}}` = m(
`D_(p)`)

`{\bar{M_(n)}}` =
`(42.08* 25000)`

`{\bar{M_(n)}}` = 1052000 g/mol

where,

`{\bar{M_(n)}}` = number-average molecular weight