Some Basic Concepts of Chemistry 1 Question 4
4. $10 \mathrm{~mL}$ of $1 \mathrm{mM}$ surfactant solution forms a monolayer covering $0.24 \mathrm{~cm}^{2}$ on a polar substrate. If the polar head is approximated as a cube, what is its edge length?
(2019 Main, 9 April II)
(a) $2.0 \mathrm{pm}$
(b) $0.1 \mathrm{~nm}$
(c) $1.0 \mathrm{pm}$
(d) $2.0 \mathrm{~nm}$
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Solution:
- Given, volume $=10 \mathrm{~mL}$
Molarity $=1 \mathrm{mM}=10^{-3} \mathrm{M}$
$\therefore$ Number of millimoles $=10 \mathrm{~mL} \times 10^{-3} \mathrm{M}=10^{-2}$
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 $=0.24 \mathrm{~cm}^{2}$
$\therefore$ Surface area occupied by 1 molecule
$$ =\frac{0.24}{6 \times 10^{18}}=0.04 \times 10^{-18} \mathrm{~cm}^{2} $$
As it is given that polar head is approximated as cube. Thus, surface area of cube $=a^{2}$, where
$$ \begin{aligned} a & =\text { edge length } \ \therefore \quad a^{2} & =4 \times 10^{-20} \mathrm{~cm}^{2} \ a & =2 \times 10^{-10} \mathrm{~cm}=2 \mathrm{pm} \end{aligned} $$
