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.

$0.10\,\mathrm{mm}^3 = 10^{-1}\,\mathrm{mm}^3$

$400.\,\mathrm{cm}^2 = 4\times10^2\,\mathrm{cm}^2$

Then

$\dfrac{10^{-1} \,\mathrm{mm}^3}{4\times10^2\,\mathrm{cm}^2} \cdot \dfrac{\left(\frac{1\,\rm m}{10^3\,\rm mm}\right)^3}{\left(\frac{1\,\rm m}{10^2\,\rm cm}\right)^2} = \dfrac{10^{-1}\times10^{-9} \,\mathrm m^3}{4\times10^2\times10^{-4}\,\mathrm m^2} = \dfrac{10^{-10}}{4\times10^{-2}}\,\mathrm m \\\\ ~~~~~~~~= 0.25\times10^{-8}\,\mathrm m$

Now, 1 m = 10⁹ nm, so

$0.25 \times10^{-8}\,\mathrm m \cdot \dfrac{10^9\,\rm nm}{1\,\rm m} = 0.25\times10^1\,\mathrm{nm} = \boxed{2.5\,\rm nm}$