Ethanol and water hygroscopic equilibrium concentration
You have a closed system with (say) 100 gallons of (say) pure ethanol and 100 gallons of moist air.  There is not enough water in the air to go into the ethanol and reach the azeotropic ratio -- not by 2 or 3 orders of magnitude (see below). 

If you know the relative humidity of the air at the time the tote was opened (and the temperature), you can find the absolute humidity (the mass % of water vapor in the air) and so estimate the maximum amount of water that went into the tote.  Since 100 gallons is about 20 times the volume of a mole of gas, assuming an absolute humidity of 10% (it's very likely way less) the answer is about 2 moles, or about 40 grams. In equilibrium (in the closed system) almost all of the water vapor will very soon go into the ethanol, so you have 100 gallons of ethanol (with a bit of toluene), plus 40 g of water.  That's 0.01%.  Whether this much water ruins your alcohol or not, you will know better than anyone.

Every time you open the tote, you're potentially letting in 40g of water.
I say "potentially" because, since the gas content is not instanty exchanged, much less water than that might go in.  But still, if you open it often, the percentage will keep going up -- asymptotically to the azeotropic ratio, but since the alcohol will also evaporate, as has been pointed out, you probably won't ever get near that ratio.