Question

$10\, mL$ of $1\, mM$ surfactant solution forms a monolayer covering $0.54\, cm ^{2}$ on a polar substrate. If the polar head is approximated as a cube, what is its edge length in $pm$ ?
[Given: $N _{ A }=6 \times 10^{23} mol ^{-1}$ ]

Answer 1

Given volume $=10\, mL$
Molarity $=1\, mM =10^{-3} M$
$\therefore $ Number of millimoles $=10\, mL \times 10^{-3} M$
$=10^{-2}$
Number of moles $=10^{-5}$
Now, number of molecules of surfactant
$=$ Number of moles $\times$ Avogadro's number
$=10^{-5} \times 6 \times 10^{23}=6 \times 10^{18}$
Surface area occupied by $6 \times 10^{18}$ molecules
$=0.54\, cm ^{2}$
$\therefore $ Surface area occupied by $1$ molecule of surfactant
$=\frac{0.54}{6 \times 10^{18}}=9 \times 10^{-20} cm ^{2}$
As it is given that polar head is approximated as cube.
Thus, surface area of cube $= a ^{2}$,
where, $a=$ edge length
$\therefore a^{2}=9 \times 10^{-20} cm ^{2}$
$a=3 \times 10^{-10} cm =3\, pm$