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Calculate the area per molecule occupied by a monolayer of n-octadecane molecules adsorbed at the air-water interface assuming a surface pressure of 20 mN/m. The molecular weight of n-octadecane is 284.54 g/mol and its molecular area in the liquid phase is 0.219 nm². Use the Langmuir adsorption isotherm equation to solve the problem.

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To calculate the area per molecule occupied by a monolayer of n-octadecane molecules adsorbed at the air-water interface, we need to use the Langmuir adsorption isotherm equation: =  k * A *   /  1 -  where  is the surface pressure  20 mN/m , k is the Boltzmann constant  1.38  10^-23  J/K , A is the molecular area in the liquid phase  0.219 nm , and  is the surface coverage  fraction of the surface covered by the adsorbed molecules .First, we need to find the value of . Rearrange the Langmuir equation to solve for : =  /  k * A +  Now, we need to convert the surface pressure from mN/m to J/m: = 20 mN/m *  1 N / 1000 mN  *  1 J / 1 N*m  = 0.02 J/mNow, we can plug in the values into the equation: = 0.02 J/m /  1.38  10^-23  J/K * 0.219  10^-18  m + 0.02 J/m   0.02 J/m /  3.02  10^-22  J/m + 0.02 J/m   0.02 J/m / 0.0200000302 J/m  0.99999849Now that we have the surface coverage, we can calculate the area per molecule occupied by a monolayer of n-octadecane molecules adsorbed at the air-water interface:A_adsorbed = A_liquid /  1 -  A_adsorbed = 0.219 nm /  1 - 0.99999849 A_adsorbed  0.219 nm / 1.51  10^-6 A_adsorbed  145.03 nmSo, the area per molecule occupied by a monolayer of n-octadecane molecules adsorbed at the air-water interface is approximately 145.03 nm.

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