5
$\begingroup$

From what I understand, the larger the molecular size, the slower the rate of diffusion and vice versa. However, after doing a lab (similar to this one) about diffusion through a cellulose membrane, my instructor said that despite substances having different molecular sizes, the rate of diffusion (thru the membrane) will be the same. Could someone further explain or did I hear incorrectly?

$\endgroup$
  • $\begingroup$ Yes, diffusion coefficient varies with molecular size, so it is difficult to imagine a mechanism whereby a membrane would precisely nullify this relationship, but could you describe your experiment in a little more detail? $\endgroup$ – Alan Boyd Jul 31 '17 at 7:55
  • $\begingroup$ @AlanBoyd mayhewbiology.com/laboratory/Diffusion%20lab.htm This is essentially what we did $\endgroup$ – bioguy Jul 31 '17 at 18:09
2
$\begingroup$

I think that your instructor was wrong.

For the type of experiment that you describe the dialysis tubing (cellulose acetate with pores) is acting as a semi-permeable membrane which can block the diffusion of a polymer (starch) but will allow the passage of a small molecule (glucose). The molecular weight cut-off of the tubing was probably 10 kilodaltons or greater, so glucose (180 daltons) would diffuse through the pores without hindrance, as would sucrose (342 daltons). But would the monosaccharide and the disaccharide move through the membrane at different rates?

This table gives values for the diffusion coefficient (D) of these two sugars as:

glucose 6.7 cm² s⁻¹ × 10⁻⁶

sucrose 5.2 cm² s⁻¹ × 10⁻⁶

The time τ to travel distance L by diffusive movement is given by τ = L²/6D

Here all that matters is that τ is inversely proportional to D, so we can estimate that glucose 'diffuses' approximately 30% faster than sucrose.

Bulk diffusion is, of course, driven by a concentration gradient: if you were to start with, say, 0.1 M each of glucose and sucrose in a dialysis tube the glucose should reach equilibrium with the surrounding liquid faster than the sucrose.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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