Question

Volume of $N _{2}$ at $NTP$ required to form a monolayer on the surface of iron catalyst is $8.15 \,mL / g$ of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies $16 \times 10^{-22} m ^{2}$ area?

• A. $16 \times 10^{-16} cm_{2}$
• B. $0.35 \,m ^{2} / g$
• C. $39 \,m ^{2} / g$
• D. $22400 \,cm ^{2}$

Answer 1

Answer: (B)

$N _{2}=8.15 \times 10^{-3} L$ of $NO _{2}$

Moles of $N _{2}=\frac{8.15 \times 10^{-3}}{22.4}=0.3638 \times 10^{-3}$

Molecules $=0.3638 \times 10^{-3} \times 6.022 \times 10^{23}=2.19 \times 10^{20}$

$\therefore $ surface area covered $=\left(16 \times 10^{-29} m ^{2}\right)\left(2.19 \times 10^{20}\right)$

$=35.072 \times 10^{-9} \,m ^{2}$

$=0.35 \,m ^{2} / g$