Question

$1 \,g$ of activated charcoal has a surface area $10^{5} m ^{2}$. If there is complete monolayer coverage and has effective surface area of $NH _{3}$ is $0.015\, nm ^{2}$, then volume of $NH _{3}$ in $cm ^{3}$ at STP that could be adsorbed on $50\, g$ of charcoal is _______.

Answer 1

Total surface area available for adsorption

$=50 \times 10^{3} m ^{2}$

Effective surface area of one $NH _{3}$ molecule

$=1.5 \times 10^{-20} m ^{2}$

$\therefore $ No. of molecules of $ NH _{3}$ adsorbed

$=\frac{50 \times 10^{3}}{1.5 \times 10^{-20}} $

$=33.3 \times 10^{17}$

Moles of $NH _{3}=\frac{33.3 \times 10^{17}}{6 \times 10^{23}}$

$=5.5 \times 10^{-6}$

$\therefore V ^{ NH_3 }=\frac{ nRT }{ P }$

$=\frac{5.5 \times 10^{-6} \times 0.082 \times 273}{1}$

$=1.23 \times 10^{-4} litre$

$=0.123 \,cm ^{3}$