1
$\begingroup$

I'm presently undergoing an Experiment for the determination of the viscosity of Ficoll-70 using Ostwald viscometer to calculate the time and a digital weighing balance to determine the weight of a specified volume of the solution at different concentration. I have reached a result but unable to correlate with any data provided by the journals or thesis on this area. It's because they all have given the value of 'Intrinsic Viscosity'. How different is the Dynamic viscosity from the Intrinsic one? Can I calculate the dynamic viscosity of a solution if I know its Intrinsic viscosity ?

$\endgroup$
1

1 Answer 1

2
$\begingroup$

Your viscometer measures the viscosity of your Ficoll-70 solution. This viscosity is made up partly from the viscosity of the water and partly from the viscosity of the Ficoll-70 dissolved within it, and obviously the viscosity you measure increases with the concentration of Ficoll-70.

Intrinsic viscosity is a rather different concept. It's really a measure of the shape of the solute at infinite dilution, or at least sufficiently dilute that you can ignore three or more body interactions. For these low dilutions it's the shape of the solute that determines the viscosity, and the intrinsic viscosity depends on this shape.

At very low concentrations the dynamic viscosity will be given by:

$$ \eta = ([\eta]\phi + 1)\eta_0 $$

where $[\eta]$ is the intrinsic viscosity, $\eta_0$ is the viscosity of the solvent (presumably water in this case) and $\phi$ is the concentration of the Ficoll-70. You'll probably find $[\eta]$ is given in units of ml/g, so you need to give $\phi$ in g/ml. You could try calculating the viscosity using this formula, but I'd be suprised if it worked for the sort of concentrations Ficoll-70 is normally used at i.e. a few percent by weight.

Later: I note that https://somapps.med.upenn.edu/pbr/portal/immune/Ficcoll_info.pdf gives the relative viscosity of Ficoll-70 as a function of concentration. You could use this to check your results.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.