Question

Fatty acids spread spontaneously on water to form a monomolecular film. A solution containing 0.10 mm^3 of a fatty acid is dropped into a tray full of water. The acid spreads on the surface to form a continuous film covering an area of 400. cm^2 . What is the average film thickness in nanometers?​

Answer 1

Divide the volume by the area. Using scientific makes things a bit cleaner.

`0.10\,\mathrm{mm}^3 = 10^(-1)\,\mathrm{mm}^3`

`400.\,\mathrm{cm}^2 = 4*10^2\,\mathrm{cm}^2`

Then

`\frac{10^(-1) \,\mathrm{mm}^3}{4*10^2\,\mathrm{cm}^2} \cdot (\left((1\,\rm m)/(10^3\,\rm mm)\right)^3)/(\left((1\,\rm m)/(10^2\,\rm cm)\right)^2) = (10^(-1)*10^(-9) \,\mathrm m^3)/(4*10^2*10^(-4)\,\mathrm m^2) = (10^(-10))/(4*10^(-2))\,\mathrm m \\\\ ~~~~~~~~= 0.25*10^(-8)\,\mathrm m`

Now, 1 m = 10⁹ nm, so

`0.25 *10^(-8)\,\mathrm m \cdot (10^9\,\rm nm)/(1\,\rm m) = 0.25*10^1\,\mathrm{nm} = \boxed{2.5\,\rm nm}`