Question

$1$ g of activated charcoal has a surface area ${10}^3{m}^2$. If complete monolayer coverage is assumed and effective surface area of $N{H}_3$ molecules is $0.129n{m}^2$, how much $N{H}_3$ in moles at STP could be adsorbed on $25$ g of charcoal (write answer as the nearest integer after multiplying with 10)?

Answer 1

Total surface area available for adsorption $=25\times {10}^3{m}^2$.

Effective surface area of one $N{H}_3$ molecule $=0.129\times {10}^{-18}{m}^2$.

$\therefore$ Number of molecules of $N{H}_3$ adsorbed $=\frac{25\times {10}^3}{0.129\times {10}^{-18}}=1.94\times {10}^{23}$.

$\therefore$ Moles of $N{H}_3=\frac{1.94\times {10}^{23}}{6\times {10}^{23}}=0.323$.

Hence, $0.323$ moles of $N{H}_3$ in $c{m}^3$ at STP could be adsorbed on $25$ g of charcoal.

So, the nearest integer after multiplying with 10 is 3.