Question

$10\text{mL}$ of $1\text{mM}$ surfactant solution forms a monolayer covering $0.24c{m}^2$ on a polar substrate. If the polar head is approximated as a cube, what is its edge length?
(Take $N_A=6\times {10}^{23}$)

• A. $2\text{pm}$
• B. $0.1\text{nm}$
• C. $1\text{pm}$
• D. $2\text{nm}$

Answer 1

The correct option is A $2\text{pm}$

Number of millimoles of surfactant $=10\times {10}^{-3}\text{mmol}={10}^{-2}\text{mmol}={10}^{-5}\text{mol}$
So, total number of surfactant molecules $={10}^{-5}\times 6\times {10}^{23}$
Hence surface area oocupied by $6\times {10}^{18}$ surfactant molecules is $0.24c{m}^2$
Therefore, surface area occupied by 1 surfactant molecule $=\frac{0.24}{6\times {10}^{18}}=0.04\times {10}^{-18}c{m}^2$

As it is given that polar head is approximated as cube, the surface area of cube = $a^2$, where $a$ is edge length.
$\therefore a^2=4\times {10}^{-20}c{m}^2$
$\Rightarrow a=2\times {10}^{-10}cm=2\text{pm}$