Question

You can obtain a rough estimate of the size of a molecule with the following simple experiment: Let a droplet of oil spread out on a fairly large but smooth water surface. The resulting "oil slick" that forms on the surface of the water will be approximately one molecule thick. Given an oil droplet with a mass of 9.00 × 10−7 kg and a density of 918 kg/m3 that spreads out to form a circle with a radius of 41.8 cm on the water surface, what is the approximate diameter of an oil molecule?

Answer 1

Answer:

The diameter of the oil molecule is $4.4674\times 10^{-8} cm$ .

Explanation:

Mass of the oil drop = $m=9.00\times 10^{-7} kg$

Density of the oil drop = $d=918 kg/m^3$

Volume of the oil drop: v

$d=\frac{m}{v}$

$v=\frac{m}{d}=\frac{9.00\times 10^{-7} kg}{918 kg/m^3}$

Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.

Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m

1 cm = 0.01 m

Surface of the sphere is given as: a = $4\pi r^2$

$a=4\times 3.14\times (0.418 m)^2=2.1945 m^2$

Volume of the oil drop = v = Area × thickness

$\frac{9.00\times 10^{-7} kg}{918 kg/m^3}=2.1945 m^2\times x$

$x= 4.4674\times 10^{-10} m= 4.4674\times 10^{-8} cm$

The thickness of the oil drop is $4.4674\times 10^{-8} cm$ and so is the diameter of the molecule.