0.27 g of a long chain fatty acid was dissolved in 100 cm$$^3$$ of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer?
[Density of fatty acid = 0.9 g cm$$^{-3}$$; $$\pi = 3$$]

We have a solution in which $$0.27\ \text{g}$$ of the long-chain fatty acid is dissolved in $$100\ \text{cm}^3$$ (that is, $$100\ \text{mL}$$) of hexane.

First we calculate the concentration of fatty acid in the solution. By definition

$$\text{Concentration}=\frac{\text{mass of solute}}{\text{volume of solution}} =\frac{0.27\ \text{g}}{100\ \text{cm}^3}=0.0027\ \text{g cm}^{-3}.$$

Now, only $$10\ \text{mL}=10\ \text{cm}^3$$ of this solution is taken. The mass of fatty acid present in this aliquot is therefore

$$m=\left(0.0027\ \text{g cm}^{-3}\right)\left(10\ \text{cm}^3\right)=0.027\ \text{g}.$$

When the hexane evaporates, this entire mass spreads out on the water surface to form a monomolecular layer. To find the thickness (height) of this layer, we first need the volume it occupies. Using the relation

$$\text{Volume}=\frac{\text{mass}}{\text{density}},$$

and the given density $$\rho =0.9\ \text{g cm}^{-3},$$ we get

$$V=\frac{0.027\ \text{g}}{0.9\ \text{g cm}^{-3}}=0.03\ \text{cm}^3.$$

Next we determine the area over which the fatty acid spreads. The watch glass has a circular water surface whose radius is the distance from edge to centre, namely $$r=10\ \text{cm}.$$ The area of a circle is given by

$$A=\pi r^{2}.$$

With the approximation $$\pi =3,$$ we have

$$A=3\,(10\ \text{cm})^{2}=3\times100\ \text{cm}^2=300\ \text{cm}^2.$$

The thickness (height) $$h$$ of the monolayer is the volume divided by the area:

$$h=\frac{V}{A}=\frac{0.03\ \text{cm}^3}{300\ \text{cm}^2}=0.0001\ \text{cm}=1\times10^{-4}\ \text{cm}.$$

To express this height in metres we use $$1\ \text{cm}=10^{-2}\ \text{m},$$ so

$$h=1\times10^{-4}\ \text{cm}\times10^{-2}\ \text{m cm}^{-1}=1\times10^{-6}\ \text{m}.$$

Hence, the correct answer is Option B.