# Liquids and surface tension

1. Oct 23, 2011

### KatrinaM

1. The problem statement, all variables and given/known data

cphenol/mole dm−3 0.05 0.08 0.127 0.268 0.496
γ/mN m−1 67.88 64.60 60.10 51.58 44.97

Estimate the relative surface excess concentrations of phenol in the concentration
ranges of (i) 0.05 to 0.08 mole/dm3 and (ii) 0.268 to 0.496 mole/dm3 at
the surface of an aqueous solution from the following data taken at 293 K
Given that the cross section of a phenol molecule is about 0.4 nm2 estimate
what fraction of the surface is covered by phenol at the higher of the two
concentration ranges.

2. Relevant equations
dγ = − ΓRTd ln c, where lambda is the surface excess concentration of phenol relative to that of water, gamma is the surface tension, and c the concentration of phenol
the surface excess concentration of component phenol equals the number of moles of surface excess quantities divided by the surface area

3. The attempt at a solution
Basically I worked out the first question, getting answers of 2.8E-6 mol/m^2 and 4.41E-6 mol m^-2, but I have little idea on how to answer the last question. I tried to work backwards, I'm guessing it has something to do with working out the number of moles of phenol per m^2 of the surface and multiplying by the surface area of a phenol molecule, but I'm not sure how to calculate that, and I'd be grateful if you give a hint in the right direction.