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Unformatted text preview: Water Octanol Here we are looking at a single solute in a (very small) fixed volume, a, of octanol or water. Because the water is more dense by a factor of 8.748, it has ~9 times as many molecules of water per unit volume (we just rounded 8.748 to 9). !
We know the solute density ratio, ! !"# , in this situation is 1, since there is just 1 !"# solute per unit volume, a, in each case. !
! !"# = 1 !"# We want to know how this relates to P, the ratio of equilibrium mole fractions. !
!/!
! = ! !"# = !/!" = 9 !"# So to scale P to get the solute density ratio, we have to divide P by 9 (remember it’s really 8.748). So that means we need to divide P by 8.748 to get the density ratio. ! !.!"# = !!"# !!"# ...
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This note was uploaded on 09/12/2011 for the course CHEM 107B taught by Professor Jamesames during the Spring '09 term at UC Davis.
 Spring '09
 JAMESAMES
 Physical chemistry, Mole, pH

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