Question

7) What is the effective area of charcoal occupied per ammonia molecule.
$$\begin{gathered} \left(A\right)\frac{1}{2}\sqrt{3}\times {10}^{-20}{m}^2 \\ \left(B\right)4\sqrt{3}\times {10}^{-20}{m}^2 \\ \left(C\right)2\sqrt{3}\times {10}^{-20}{m}^2 \\ \left(D\right)8\sqrt{3}\times {10}^{-20}{m}^2 \end{gathered}$$

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Surafce}\mathrm{are}\mathrm{of}1\mathrm{g}\mathrm{of}\mathrm{activated}\mathrm{charcoal}={10}^3{\mathrm{m}}^2 \\ \mathrm{Therefore},\mathrm{surafce}\mathrm{area}\mathrm{of}10\sqrt{3}\mathrm{g}\mathrm{of}\mathrm{activated}\mathrm{charcoal}=10\sqrt{3}\times {10}^3={10}^4\sqrt{3} \\ \mathrm{Now},{\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23} \\ \mathrm{Moles}\mathrm{of}{\mathrm{NH}}_3=5\mathrm{M}\times 0.1\mathrm{L}=0.5\mathrm{mol} \\ \mathrm{Hence},\mathrm{effective}\mathrm{area}=\frac{0.5\times {10}^4\times \sqrt{3}}{6\times {10}^{23}}=8\sqrt{3}\times {10}^{-20}{\mathrm{m}}^2 \end{gathered}$$