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 the value of Avogadro's number is_______.

• A. $3\times {10}^{23}$
• B. $6\times {10}^{23}$
• C. $4\times {10}^{23}$
• D. $1\times {10}^{23}$

Answer 1

The correct option is A $3\times {10}^{23}$

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

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

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

So, Avogadro's number is $3.30\times {10}^{23}$

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