This may be more appropriately listed as a math question...  It could
go either way - math/chemistry.

Question: If you were to generate a CMC (critical micelle
concentration) plot of a surfactant in water, you would have a plot of
surface tension (y-axis) versus the log of the surfactant
concentration (x-axis).  These are common in surface science
literature.  According to theoretical calculations (discussed below),
you can use the slope of this plot (where surface tension is lowering)
to calculate the area/molecule at the air/water interface so long as
you measured your surfactant concentration in M (moles/liter).  IF you
measured your surfactant concentration in ppm (parts per million)
instead of M, can you still take the slope of the CMC plot and plug it
into the same equations to determine surface area per molecule?  If
not, what/how would you do the conversion to determine the
area/molecule from a plot using ppm rather than M (eg. multiply or
divide by the molecular weight of the surfactant, etc.?)

Background/Reference Material:
Surface Area Per Molecule is a common calculation in the surface
science world.  It is typically recorded in "square angstroms per
molecule."  You can reference Surfactants and Interfacial Phenomena
Third Edition (by Milton J. Rosen), in chapter 2 section III.B to see
how this calculation is being made.  Of course, you can likely find
this calculation many other places...  it is fairly common in the
surface science world.

In order to calculate area/molecule, it has been shown (common
literature) that you can use the slope of a plot of surface tension
(y-axis) versus log of surfactant concentration (x-axis) to calculate
the "surface excess concentration."  Then the Surface Excess
Concentration can be plugged into an equation w/ other constants to
give the Area/Molecule in square angstroms/molecule.  (I'm sure you'll
find these calculations fairly early/easy in your research, so I'll
not take the time to type them all out here...)

All of the examples that I have seen of this calculation depend on a
plot of surface tension versus log surfactant concentration, in which
the surfactant concentration is recorded M (moles/L).  This makes
sense since the equations that I'm referencing end up using the gas
constant and Avogadro's number to ultimately calculate area per
molecule.  However, when you calculate Surface Excess Concentration,
the value that you plug in from your plot of surface tension versus
log concentration (molar concentration) is simply a slope - delta y
divided by delta x.  Therefore, is it equivalent (will I get the same
answer) to plug in the slope of a plot of surface tension vs log
surfactant concentration where my surfactant concentration is recorded
in ppm rather than in M?

Said another way: Wouldn't the slope of a plot of surface tension
versus surfactant concentration (X axis on a log scale) be the same if
surfactant concentration were recorded in moles/Liter and/or in ppm?

Can you please explain the math/rationale for how/why these are
different?  If they are different, is there a way to use the surface
tension/concentration (ppm) plot/slope to convert/calculate
area/molecule by multiplying/dividing by molecular weight of the
surfactant?

Feel free to ask me for more/clear details.  I am operating under the
assumption that you can fairly readily find info. on the following
topics:

critical micelle concentration plots
calculation of surface concentration and surface excess concentrations
area per molecule at an interface
the Gibbs Adsorption Equation

Many Thanks,
WMXO

Basically, you need to convert from a concentration in units of ppm
(micrograms of solvent per gram of solution) to units of Molarity
(moles of solute per liter of solution).  To do this conversion, you
need to know how the density of the solution varies as a function of
the solute concentration in ppm.  To see why this is so, let X be the
conversion factor between ppm and Molar units:

   ppm * X = Molar
    
    (grams solute)*10^6/(grams solution) * X = (moles solute)/(liter solution)

   X = (moles solute)/(liter solution)*(grams solution)/[10^6*(grams solute)]
   X = 10^-6*[(moles solute)/(grams solute)]*[(grams solution)/(liter solution)]
   X = 10^-6*(1/(molecular weight solute)]*[density solution]

where the molecular weight is expressed in g/mol, and the density is
expressed in grams/liter.

For an ideal solution, the volume of the solution is simply given by
the sum of the molar volumes of solvent and solute multiplied by the
number of moles of each present in the solution.  Most solutions are
not ideal, though.  You can add a substantial amount of common salt to
water without changing the volume of the solution appreciably.  Such a
solution has a negative volume of mixing (the solution is denser than
one might expect from simply considering the volumes of the salt and
water one started with), and the solution's constituents (i.e.,
molecules and ions) of the solution are packed more closely together
than they would be in an ideal solution.  This also means that the
surface density of the constituents will be higher (i.e., the surface
area per specie is less) than if the solution were ideal.

Other substances can have a positive volume of mixing, in which case,
the volume of the solution is larger than that of the weighted sum of
the molar volumes of solvent and solute.