Question

$H_2$ gas was adsorbed on $1$ g powdered copper surface forming monolayer of molecules. On desorption, total $H_2$ collected measured $1.36c{m}^2$ at STP. Assuming the volume of $1$ molecule of $H_2$ to be $4.472\times {10}^{-23}c{m}^3$, the specific area of copper powder turns out to be $y\times {10}^{2}c{m}^2$. Find the value of $y$.

Answer 1

Number of molecules of $H_2$ in $1.36c{m}^3$ $=\frac{6.023\times {10}^{23}\times 1.36}{22400}=3.66\times {10}^{19}$

$\therefore$ Volume of $H_2$ molecule $=4.742\times {10}^{-23}$

$\therefore \frac{4}{3}\pi r^3=4.742\times {10}^{-23}$

$\therefore r=2.245\times {10}^{-8}cm$

Area of cross-section of $H_2$ molecule $=\pi r^2=3.14\times (2.245\times {10}^{-8}{)}^2$ $=1.583\times {10}^{-15}c{m}^2$

Area of molecules adsorbed $=(3.66\times {10}^{19})\times 1.583\times {10}^{-15}c{m}^2=$ Area of adsorption of $Cu$ powder $=$ Specific area of $Cu$ powder

$\therefore$ Specific area of adsorption of $Cu$ $=1.583\times {10}^{-15}\times 3.66\times {10}^{19}=579\times {10}^{2}c{m}^2$

So, the value of $y$ is $579$.