Question

When stearic acid $\left(C_{18}{H}_{36}{O}_2\right)$ is added to water, its molecules collect at the surface and form a monolayer. The cross-sectional area of each stearic acid molecule is $0.21n{m}^2$. If $1.4\times {10}^{-4}g$ of stearic acid is needed to form a monolayer over water in a dish of diameter $20cm$. (The area of the circle of radius $r$ is ($\pi r^2$), then what is equivalent of $1g$ $H$ in amu for this value of Avogadro's number?

• A. $1.66\times {10}^{-24}g$
• B. $3.33\times {10}^{-24}g$
• C. $2.5\times {10}^{-24}g$
• D. $1\times {10}^{-23}g$

Answer 1

The correct option is B $3.33\times {10}^{-24}g$

The area of the dish of $20cm$ diameter ($10cm=0.1m$ radius) has area $=\pi r^2=\pi \times 0.01m^2$

$0.21\times {10}^{-18}{m}^2$ is the area of $1$ molecule

$\therefore \pi \times 0.01m^2$ is the area of $\frac{\pi \times 0.01}{0.21\times {10}^{-18}}$ molecules

$1.4\times {10}^{-4}g$ stearic acid$=\frac{1.4\times {10}^{-4}}{284}$ mol

$=\frac{1.4\times {10}^{-4}}{284}{N}_o$ molecules
$=\frac{1.4\times {10}^{-4}{N}_o}{284}$ = $\frac{\pi \times 0.01}{0.21\times {10}^{-18}}$

$N_o=3.30\times {10}^{23}$

1 amu is $\frac{1}{N_o}=\frac{1}{3.03\times {10}^{23}}=3.33\times {10}^{-24}g$.

Hence, the correct option is $\left(B\right)$