Question

Measurements on two essentially monodisperse fraction of a linear polymer, A and B, yield molecular weights of 100,000 and 400,000, respectively. Mixture 1 is prepared from one part by weight A and two parts by weight B. Mixture 2 contains two parts by weight of A and one of B. Determine the weight-average and number-average molecular weight average of mixtures 1 and 2.

Answer 1

Answer:

Explanation:

weight average = fraction of A X MW of A + fraction of B x MW of B  

Mixture 1  

1 / 3 x 100000 + 2/3 x 400000 = 900000 / 3 = 300000

Mixture 2

2/3 x 100000 + 1/3 x 400000 = 600000 / 3 = 200000

Number average  

Mixture 1  

(1 / 100000 x 100000 + 2 / 400000 x 400000) / (1/100000 + 2/400000 )

= 3 x 400000 / 6  

= 200000

Mixture 2  

(2/100000 x 100000 + 1/400000 x 400000 )/ (2/100000 + 1 / 400000 )

3 x 400000 / 9  

=( 4/3 ) x 100000