Stearic acid, C17H35COOH has a density of 0.85 g/cm3. The molecule occupies an area of 0.205 nm2 in a closed packed surface film. Calculate the length of the molecule.

Question

Stearic acid, $\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COOH}$ has a density of $0.85 \mathrm{~g} / \mathrm{cm}^3$. The molecule occupies an area of $0.205 \mathrm{~nm}^2$ in a closed packed surface film. Calculate the length of the molecule.

(A) $3.8073 \mathrm{~nm}$

(B) $8.3073 \mathrm{~nm}$

(C) $2.7073 \mathrm{~nm}$

(D) $1.8073 \mathrm{~nm}$

Answer 1

Molar mass of stearic acid $=284$

$\text { Mass of one molecule }=\frac{284}{6.02 \times 10^{23}} \mathrm{~g}$

$\begin{array}{r}\text { Volume of one molecule }=\frac{284}{6.02 \times 10^{23} \times 0.85} \\=55.5 \times 10^{-23} \mathrm{~cm}^3\end{array}$

Area occupied by the molecule $=0.205 \mathrm{~nm}^2$

$\begin{array}{l}\quad=0.205 \times 10^{-18} \mathrm{~m}^2=0.205 \times 10^{-14} \mathrm{~cm}^2 \\\text { Hence, length of the molecule }=\frac{\text { volume }}{\text { area }} \\=\frac{55.5 \times 10^{-23}}{0.205 \times 10^{-14}}=270.73 \times 10^{-9}\mathrm{~cm}=2.7073 \mathrm{~nm}\end{array}$

So, The correct option is (C).