One litre vessel having $20g$ charcoal ( density = $2.0g/c{m^3}$ ) , was filled with a gas at $300K$ . The pressure of a gas was $760torr$ . Due to adsorption , the pressure of gas falls to $608torr$ . What is the number of gas molecules adsorbed per g of charcoal ?
A.$2.4 \times {10^{21}}$
B.$2.4 \times {10^{20}}$
C.$2.4 \times {10^{22}}$
D.$2.4 \times {10^{23}}$

Hint: We will solve this question with the help of an ideal gas equation . Also adsorption is the phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid or solid resulting in a higher concentration of the molecules on the surface .

Complete answer:
Ideal gas equation is an equation which gives the simultaneous effect of pressure and temperature on the volume of a gas .
The mathematical expression of ideal gas equation is
$PV = nRT$
where , P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = universal gas constant
T = temperature of the gas
It is given that the initial pressure is $760torr$ but then it falls to $608torr$
So , the change in pressure is
$\Delta P = 760 - 608 = 152torr$
Now we have to convert the pressure into atm to use it in the ideal gas equation .
$P = \dfrac{{152torr}}{{760torr/atm}} = 0.2atm$
We have been given the volume = 1 litre , temperature = $300K$ , R = $0.0821Latm{K^{ - 1}}mo{l^{ - 1}}$
On substituting the values in the ideal gas equation , we get
$\Delta n = \dfrac{{PV}}{{RT}}$
$ \Rightarrow \Delta n = \dfrac{{0.2 \times 1}}{{0.0821 \times 300}} = 8.12 \times {10^{ - 3}}mol$
Therefore , the number of gas molecules adsorbed = $8.12 \times {10^{ - 3}} \times {N_a}$
where N = avogadro's number
Therefore , total number of molecules = $8.12 \times {10^{ - 3}} \times 6.023 \times {10^{23}} = 4.89 \times {10^{21}}$
Now , we have to find out the number of gas molecules adsorbed per g of charcoal
$ \Rightarrow n = \dfrac{{4.89 \times {{10}^{21}}}}{{20g}} = 2.4 \times {10^{20}}molecules$
Hence option B is correct .

Note:
The universal gas constant has various values according to the units of pressure and volume , so we should choose the one accordingly . Like for example in the given question we had pressure in atm and volume in litre so we took the value of R as $0.0821Latm{K^{ - 1}}mo{l^{ - 1}}$ .