mass are available,nandn^0 , and thus the length of a statistical chain element
b, can be established. These data can then be used to predict behavior. To
take an example, the hydrodynamic radiusrhof an unbranched polymer
molecule is to a good approximation given by
rh¼2
3
rg ð 6 : 5 ÞThis allows calculation of its diffusion coefficient according to Eq. (5.10), or
of its sedimentation rate in a centrifugal field. It is also of interest in relation
to the effect of the polymer on viscosity, which is discussed in the following
section.
Question
Consider a solution of amylose in 0.33 molar KCl. In this condition, the excluded
volume parameter b¼0. Assume M¼ 106 Da. What would be the diffusion
coefficient of an amylose molecule in a very dilute solution? And what amylose
concentration would be needed for the amylose molecules plus entrapped solvent to
fill the whole volume?
Answer
Amylose is essentially a linear chain of anhydroglucose monomers. Molar mass of
the monomer thus equals that of glucose minus water, i.e., 162 Da. This means that
n¼ 106 = 162 ¼6173 monomers. From tabulated bond lengths and angles it can be
calculated that the monomer lengthL& 0 :5 nm. Table 6.1 givesb=L&5, leading to
b& 2 :5 nm andn^0 ¼nL=b&1235. To findrg, Eq. (6.4) can be applied withn¼ 0 : 5
(because the solvent is ideal), yielding 36 nm. According to Eq. (6.5), the hydro-
dynamic radius would be 2/3 timesrg, yielding 24 nm. Applying Eq. (5.16) with
T¼300 K andZ¼ 10 ^3 Pa?s yields a diffusion coefficientD& 9? 10 ^12 m^2 ?s^1 ,
which is by a factor 75 smaller thanDfor glucose. It may be noticed that the simple
assumption of a molecular radius proportional toM^1 =^3 , as applies for compact
molecules, would lead to a diffusion coefficient smaller than that of glucose by a
factorð 106 = 180 Þ^1 =^3 &18, rather than 75.
To calculate the volume occupied, we need to have a reasonable value for the
effective radius. Figure 6.5 shows that it would be roughly 1.5 timesrgfor an ideal
solvent, i.e., 54 nm. The volume of a molecule isð 4 = 3 Þpr^3 , leading to 6: 6? 10 ^22 m^3.
This implies that 15? 1020 molecules can be packed in a m^3 and dividing byNAVleads
to 0.0025 moles. Multiplying byMgives 2:5kg?m^3 , or about 0.25%w/w. A very
small concentration would thus be sufficient to ‘‘fill’’ the whole system.