Question

The volume of $N_2$ at STP required to form a mono layer on the surface of an iron catalyst is $22.4\text{mL/g}$ of the adsorbent. What will be the surface area of the adsorbent per $g$ if each nitrogen molecule occupies $16\times {10}^{-22}{\text{m}}^2$?

• A. $89600{\text{cm}}^2$
• B. $0.96{\text{m}}^2$
• C. $93{\text{m}}^2$
• D. $22400{\text{cm}}^2$

Answer 1

The correct option is B $0.96{\text{m}}^2$

The volume of $N_2$ at STP required to cover the iron surface with monolayer = $22.4{\text{mL g}}^{-1}$.
Area occupied by a single molecule $=16\times {10}^{-18}{\text{cm}}^2$
We know that,
$22400\text{mL}$ of $N_2$ at STP contains $N_A$ molecules of $N_2$

$\therefore 22.4$ mL of $N_2$ contains
$=\frac{22.4\times N_A}{22400}$
$=6.022\times {10}^{20}$ molecules of $N_2$.

Area occupied by $6.022\times {10}^{20}$ molecule of $N_2$
$=6.022\times {10}^{20}\times 16\times {10}^{-18}{\text{cm}}^2$
$=96.352\times {10}^2{\text{cm}}^2$
$=0.96{\text{m}}^2$