Question

Volume of ${N_2}$ at NTP to form a monolayer on the surface of the iron catalyst is ${\text{8}}{\text{.15ml/gram}}$ of the absorbent. What will be the surface area of the adsorbent per gm if each nitrogen molecules occupies ${\text{16}} \times {\text{1}}{{\text{0}}^{{\text{ - 22}}}}{{\text{m}}^{\text{2}}}$
A.${\text{16}} \times {\text{1}}{{\text{0}}^{{\text{ - 22}}}}{\text{c}}{{\text{m}}^{\text{2}}}$
B.$3.5 \times {10^{ - 4}}{\text{c}}{{\text{m}}^{\text{2}}}$
C.$39{{\text{m}}^{\text{2}}}/{\text{g}}$
D.$22400{\text{c}}{{\text{m}}^{\text{2}}}$

Answer 1

To answer this question you should know the concept of the ideal gas equation. Use the ideal gas equation to find the number of moles of the gas adsorbed. Use this value to calculate the total surface area covered by these molecules.
The formula used:
${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}}$
$PV = nRT$
where $P$ is pressure, $V$ is volume, $R$ is the universal gas constant, $n$ is no. of moles and $T$ is temperature

Complete step by step answer:
First, we have to calculate the moles of ${N_2}$ at NTP. We have the value of
P$$${\text{ = 1 atm}}$$; $$V{\text{ = 8}}{\text{.15ml/gram = 8}}{\text{.15}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}}{\text{L/g}}$$ T = {\text{ }}273{\text{ K}};$$R = 0.0821{\text{ L atm/ mole K}}$$. Now substituting these values in the ideal gas equation: n = \dfrac{{{\text{1 atm}} \times {\text{8}}{\text{.15}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}}{\text{L/g}}}}{{0.0821{\text{ L atm/ mole K}} \times 273{\text{ K}}}}$Solving this we will get the value of$n = 3.64 \times {10^{ - 4}}{\text{mole/g}}. Now we have to calculate the number of molecules of $${N_2}$$ required: As, 1 mole of $${N_2}$$ has 6.022 \times {10^{23}}$number of molecules so,$3.64 \times {10^{ - 4}}{\text{mole}}$per gram will contain$3.64 \times {10^{ - 4}} \times 6.022 \times {10^{23}} = 2.19 \times {10^{20}}$number of molecules per gram Now we have to calculate the total surface area of absorbent: Total surface area of absorbent:$(2.19 \times {10^{20}}) \times ({\text{16}} \times {\text{1}}{{\text{0}}^{{\text{ - 22}}}}{{\text{m}}^{\text{2}}}) = 0.35{{\text{m}}^{\text{2}}}{\text{/g}}$.
Therefore, the total surface area of absorbent per gram is $0.35{{\text{m}}^{\text{2}}} = 3.5 \times {10^{ - 4}}{\text{c}}{{\text{m}}^{\text{2}}}$.

Hence, the correct answer to this question is option B.

Note:
We shall assume ideal gas unless mentioned otherwise in the question. The reaction of gases mostly takes place by their adsorption on a metal catalyst, for ex-iron. The metal preferred is mostly a d-block element due to its ability to exist in variable oxidation states. The catalyst and the reactants are in different phases and this process is also known as heterogeneous catalysis.