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1) when n-hexatriacontanoic acid, CH3(CH2)34COOH was placed in water, a complete monomolecular film was formed. if the length of a carbon-carbon single bond in the chain is 1.54 amstrong, and each bond makes an angle of 35degrees with the vertical, calculate the effective cross sectional area of a molecule of the acid in the film. density of the film is 0.87g/cm3.

for this question, i managed to find out the length of each carbon-carbon single bond via trigo formula, and from there, i calculated the length of the hydrocarbon chain standing upright above the film, which was found to be 44.1 armstrong. however, im not sure how to make use of the density to find out the volume, as i would like to use the formula, vol/area=length of alkyl chain to obtain the cross sectional area.

2) part 1)

the adsorption of a non-ionic surface-active agent at the air surface of an aqueous solution obeys the gibbs equation, surface excess=-dy/(RTdlnc)

the values of the surface tension of an aqueous solution of a soluble polymeric surfactant at concentrations just below its critical micelle concentrations (10-3M) at 300K is are given as below.

Concentration/M Surface tension/mNm-1

5x10-5 37.4

1x10-4 36.5

5x10-4 34.4

1x10-3 33.9

use these data to calculate the surface excess of the surfactant and hence calculate the area per molecule at the air/water interface.

part 2) the hydrophilic part of polymeric surfactant in part 1 consists of a number of ethylene oxide (EO) segments. The area per molecule occupied by the small molecule surfactant C12H25(OC2H4)6OH just below its critical micelle concentration is 0.55 x 10-18 m2. Assuming that the area per EO segment is the same for the two surfactants in their saturated monolayers, estimate the number of EO groups in the polymeric surfactant.

Thanks!