But... the 99% starts out with less water in it. So even if the alcohol evaporates more quickly, it reaches dynamic equilibrium at 91% and continues to evaporate at 91%, but it has reached that point more quickly.
I don't have a citation. And my chemistry is pretty poor.
This is from wikipedia:
The azeotrope is at a mole fraction of about 0.68. The molecular weights of isopropanol and water are 60.1 and 18.0 g/mol respectively. So the weight fraction of isopropanol at the azeotrope is approximately:
(0.68*60.1) / (0.68*60.1 + (1-0.68)*18.0) = 0.88
Converting that to volume is more complicated because there is contraction upon mixing. But if you ignore that you can get sort of close. The room temperature densities of isopropanol and water are 0.79 and 1.00g/cm3, so:
(0.88/0.79) /(0.88/0.79 + (1-0.88)/1.00) = 0.90
My seat of the pants calculation puts the azeotrope at 90% v/v. But that can't be right since the drug store sells isopropanol/water at 91%. The volume contraction or my inaccurate read of the chart in wikipedia have thrown off the value slightly.
I'm surprised it's so hard to find an online reference to the v/v percentage. But here's one.
You can't distill past the azeotrope using normal methods. But if you manage to get to a higher concentration by other means then the solution will have a vapor that is enriched in the less volatile component. That is, the water will vaporize faster than the alcohol in a 99% isoprop. soln.
But the azeotrope is a boiling point minimum; hence a vapor pressure maximum. So evaporation rate would be higher for the less concentrated stuff. I'm not convinced this matters that much. And eyewitness accounts of evaporation rate are subject to bias.
All that said, it could well be that pure isopropyl alcohol is superior to a 91% mixture, I really don't know. I wouldn't be surprised if some other solvent would be better overall for cleaning the gunk off of PCBs.