A new polymer has [h] = 5.5 cm3/g, and an elution volume of 160 cm3. Based on th

Question

A new polymer has [h] = 5.5 cm3/g, and an elution volume of 160 cm3. Based on the method of Figure 3.23, what is its molecular weight?

Figure 3.23 The universal calibration curves for polystyrene and poly(vinyl acetate) (94). The number 5 in the x-axis units means that the scale is in siphon "counts" of 5 cm, so that the x-ordinate 30 corresponds to an elution volume of 150 cm3. (R. Dietz, private communication, November 1984.) Mr is the "peak" GPC molecular weight, usually the unknown, M, values are close to the geometric mean of Mn and My 10 10 103 104 25 30 35 Sem3

Explanation / Answer

GPC is used to calculate the molecular weight of polymers as well as PDI (poly dispersivity index)

the relation between molecular weight and elution volume is given by

log M= b-cVe

M- molecular weight

b,c- constants

Ve- elution volume

Using the graph, make two equations

M corresponding to elution volume 150 is 106 and for 108 we have elution volume 130

so,

log 106= b-c150 => 6=b-c150

log 108= b-c130 => 8=b-c130

solving these two you will get values b=21 and c=1/10

now for the unknown polymer Ve= 160cm3

putting in the equation

logM= b-cVe

logM= 21- (160/10)

logM= 21-16

logM= 5

taking antilog on both sides,

M= 105

So, the molecular weight of unknown polymer is 105

note: you can also cross check by comparing with the graph,105 corresponds to 160 (32).

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