
1 g of activated charcoal has a surface area of ${10^3}{m^2}$. 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 the surface of 25 g of the charcoal (write the answer as the nearest integer after multiplying it with 10)?
Answer
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Hint: First, we will find the total surface area available for adsorption, since the surface area and mass of charcoal is given in the question. Then, we will find the number of molecules of $N{H_3}$ adsorbed by dividing the total surface area by the given effective area of one $N{H_3}$ molecule. And if we know the number of molecules of $N{H_3}$, we can find the moles of $N{H_3}$.
Complete step by step answer:
Given in the question,
Weight of activated charcoal = 1 g
Surface area of activated charcoal $ = {10^3}{m^2}$
Weight of charcoal to be adsorbed by $N{H_3} = 25g$
$ \Rightarrow $ Total surface area available for adsorption $ = 25 \times {10^3}{m^2}$
Effective surface area of one $N{H_3}$ molecule $ = 0.129n{m^2}$
Effective surface area of one $N{H_3}$ molecule $ = 0.129 \times {10^{ - 18}}{m^2}$
(Since, 1 nm = ${10^{ - 9}}$ m, so $1n{m^2} = {10^{ - 18}}{m^2}$ )
$ \Rightarrow $ The number of molecules of $N{H_3}$ adsorbed $ = \dfrac{{25 \times
{{10}^8}}}{{0.129 \times {{10}^{ - 18}}}} = 193.79 \times {10^{21}} = 1.94 \times {10^{23}}$
$ \Rightarrow $ Moles of $N{H_3} = 0.323 \times \dfrac{{1.94 \times {{10}^{23}}}}{{6.022 \times
{{10}^{23}}}} = 0.323$ (Since, Avogadro’s number \[ = 6.022 \times {10^{23}}\] )
Hence, 0.323 moles of $N{H_3}$ in $c{m^3}$ at STP could be adsorbed on 25 g of charcoal.
Therefore, the nearest integer after multiplying with 10 is 3.
Note: A material’s surface characteristics refer to the properties that are associated with its surface. Typically, the measurements of surface area, surface roughness, pore size, and reflectivity constitute the surface characteristics. Information related to surface characteristics is of utmost importance for the possible applications of surfaces as semiconductors, heterogeneous catalysts and also in biological research.
Complete step by step answer:
Given in the question,
Weight of activated charcoal = 1 g
Surface area of activated charcoal $ = {10^3}{m^2}$
Weight of charcoal to be adsorbed by $N{H_3} = 25g$
$ \Rightarrow $ Total surface area available for adsorption $ = 25 \times {10^3}{m^2}$
Effective surface area of one $N{H_3}$ molecule $ = 0.129n{m^2}$
Effective surface area of one $N{H_3}$ molecule $ = 0.129 \times {10^{ - 18}}{m^2}$
(Since, 1 nm = ${10^{ - 9}}$ m, so $1n{m^2} = {10^{ - 18}}{m^2}$ )
$ \Rightarrow $ The number of molecules of $N{H_3}$ adsorbed $ = \dfrac{{25 \times
{{10}^8}}}{{0.129 \times {{10}^{ - 18}}}} = 193.79 \times {10^{21}} = 1.94 \times {10^{23}}$
$ \Rightarrow $ Moles of $N{H_3} = 0.323 \times \dfrac{{1.94 \times {{10}^{23}}}}{{6.022 \times
{{10}^{23}}}} = 0.323$ (Since, Avogadro’s number \[ = 6.022 \times {10^{23}}\] )
Hence, 0.323 moles of $N{H_3}$ in $c{m^3}$ at STP could be adsorbed on 25 g of charcoal.
Therefore, the nearest integer after multiplying with 10 is 3.
Note: A material’s surface characteristics refer to the properties that are associated with its surface. Typically, the measurements of surface area, surface roughness, pore size, and reflectivity constitute the surface characteristics. Information related to surface characteristics is of utmost importance for the possible applications of surfaces as semiconductors, heterogeneous catalysts and also in biological research.
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