Question

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

Answer 1

Answer: 3

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