Question

$10\text{mL}$ of one millimolar surfactant solution forms a monolayer, covering $0.24{\text{cm}}^2$ on a polar substrate. If the polar head is approximated as a cube, what is its edge length in pm?
Take $N_A=6\times {10}^{23}$

Answer 1

Number of moles of the surfactant that form a monolayer:
$n=V\times M=0.010\text{L}\times 1\text{mmol/L}=0.01\text{mmol}={10}^{-5}\text{mol}$
1 mol has $6\times {10}^{23}$ particles.
Number of molecules in the given solution is:
$={10}^{-5}\times (6\times {10}^{23})$
$=6\times {10}^{18}\text{molecules}$
Since the polar head is approximated as a cube, the area covered by a single molecule that forms a monolayer is = $a^2$, $a$ is the edge length of the cube.
The total area covered is $6\times {10}^{-18}{a}^2$ using with we can find the edge length:
$0.96{\text{cm}}^2=6\times {10}^{18}\times a^2$
$\Rightarrow a^2=\frac{0.96{\text{cm}}^2}{6\times {10}^{18}}$
$\Rightarrow a=\sqrt{16}\times {10}^{-10}\text{cm}$
$=4\text{pm}$