Question

One gram of activated charcoal has a total surface area equal to${10^3}{m^2}$. If the radius of gaseous molecules is ${10^{ - 8}}cm$ and the gas shows only monolayer adsorption on the surface of the charcoal, then the volume of gas at STP adsorbed on the total surface of $\dfrac{{22}}{7}g$ charcoal is (in liters):
A. $3.73$
B. $7.46$
C. $22.4$
D. $24.5$

Answer 1

The above question has given the total surface area of one gram of activated charcoal. So if we find the surface area required for $\dfrac{{22}}{7}g$ charcoal for the given radius then we can easily get the total number of molecules adsorbed on the charcoal surface. Then, for the calculated number of moles, we can easily calculate the volume at STP.

Formula Used: The volume of gas in liters, $V = {V_{STP}} \times n$ where $V$ represents the volume of gas at STP absorbed on the total surface, ${V_{STP}}$ represents the volume at standard temperature and pressure (STP) and $n$ represents the number of moles absorbed.

Complete step by step answer:
First, we will understand about activated charcoal. Generally, it is nothing but processed or activated carbon. It has small pores with low volume which increases the surface area available for the reaction. Now we will extract all the given quantities from the problem and then we will proceed further to calculate the number of moles absorbed.
So we have the total surface area of $1g$ activated charcoal equal to ${10^3}{m^2}$. So the total surface area for $\dfrac{{22}}{7}g$ charcoal will be $\dfrac{{22}}{7} \times {10^3}{m^2} = \dfrac{{22}}{7} \times {10^7}c{m^2}$. Now we have the radius of a single gaseous molecule as ${10^{ - 8}}cm$. Therefore, the total surface area occupied by one molecule on the surface of activated charcoal is given by $\pi {r^2} = \dfrac{{22}}{7} \times {10^{ - 16}}c{m^2}$. Now if we divide the total surface area occupied by the given amount of charcoal by the total available surface area we can get the total number of gaseous molecules absorbed on the surface.
Several gaseous molecules are adsorbed $= \dfrac{{\dfrac{{22}}{7} \times {{10}^7}}}{{\dfrac{{22}}{7} \times {{10}^{ - 16}}}} = {10^{23}}$.
Now we will calculate the number of moles of gas absorbed as $n = \dfrac{{{{10}^{23}}}}{{6 \times {{10}^{23}}}} = \dfrac{1}{6}mole$. The number of moles is the ratio of the number of molecules absorbed to the Avogadro number. Now we have $n = \dfrac{1}{6},{V_{STP}} = 22.4L$ we can easily calculate the volume of gas at STP absorbed on the total surface using the formula $V = {V_{STP}} \times n$. So now we will substitute the values,
$V = {V_{STP}} \times n = 22.4 \times \dfrac{1}{6} = 3.73L$

So, the correct answer is Option A.

Note: In monolayer adsorption, all the molecules are in contact with the total surface layer of absorbent. If there is more than one layer of adsorption it is called multilayer adsorption in which all the adsorbed molecules are in contact with the surface layer of the adsorbent.