Some Basic Concepts of Chemistry - Result Question 66
####4. $10 mL$ of $1 mM$ surfactant solution forms a monolayer covering $0.24 cm^{2}$ on a polar substrate. If the polar head is approximated as a cube, what is its edge length?
(2019 Main, 9 April II)
(a) $2.0 pm$
(b) $0.1 nm$
(c) $1.0 pm$
(d) $2.0 nm$
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Solution:
- 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
$$ \begin{aligned} & =\text { Number of moles } \times \text { Avogadro’s number } \ & =10^{-5} \times 6 \times 10^{23}=6 \times 10^{18} \end{aligned} $$
Surface area occupied by $6 \times 10^{18}$ molecules $=0.24 cm^{2}$
$\therefore$ Surface area occupied by 1 molecule
$$ =\frac{0.24}{6 \times 10^{18}}=0.04 \times 10^{-18} cm^{2} $$
As it is given that polar head is approximated as cube. Thus, surface area of cube $=a^{2}$, where
$$ \begin{aligned} a & =\text { edge length } \ \therefore \quad a^{2} & =4 \times 10^{-20} cm^{2} \ a & =2 \times 10^{-10} cm=2 pm \end{aligned} $$
