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**Hint:**We can use the ideal gas equation to determine the value of number density. Number density is the number of particles per unit volume. Hence, replacing the ratio of number of moles per unit volume with number density, we can solve this question.

**Formula used:**

$\therefore N = \dfrac{P}{{RT}} \times {N_A}$

where; $N = Number\,density,{\rho _{Mass}} = Mass\,density,{N_A} = Avogadro's\,number$

$P = pressure,T = Temperature,R = Gas\,Constant$

**Complete step by step solution:**

Let us first understand what are the terms in the question:

Number density can be defined as the degree of concentration of molecules, atoms or particle in

$3 - $Dimensional space. The general formula of number density:

$N = \dfrac{n}{V}$ where $N = Number\,density,\,n = number\,of\,particles,V = Volume$

The formula to calculate number density for gases is:

$N = {\rho _{Mass}} \times {N_A}$

Here, ${\rho _{Mass}}$= Mass Density which is ${\rho _{mass}} = \dfrac{{Mass}}{{Volume}}$ and ${N_A}$is Avogadro’s number.

We will need to Modify the value for ${\rho _{Mass}}$ since in the formular we need the value of mass, but it is very difficult to determine the mass of a gas and hence we will replace it with another variable which can be easily determined for gases.

$\therefore {\rho _{mass}} = \dfrac{P}{{RT}}$

Where, $P = pressure,T = Temperature,R = Gas\,Cons\tan t$

Now we have all the possible terms we need to calculate the number density for a gas

$P = 1bar,T = 300K,R = 8.314Jmo{l^ - }{K^ - },{N_A} = 6.022 \times {10^{23}}moles$

Substituting these values in the above equation we get:

$N = \dfrac{{1bar}}{{8.314Jmo{l^ - }{K^ - } \times 300K}} \times 6.022 \times {10^{23}}moles$

Solving the above equation, we get:

$N = 2.4 \times {10^{25}}\,{m^{ - 3}}$

**Hence, Option A is correct.**

**Additional Information:**We assume that when gas molecules collide they do not lose any energy and hence their internal energy remains constant.

**Note:**

When we substitute the value of Mass density ${\rho _{Mass}}$ as ${\rho _{mass}} = \dfrac{P}{{RT}}$ , we are assuming that the gas under consideration is an ideal gas and they have negligible volume. If these assumptions are removed, we can no longer ignore the Volume factor for real gases and hence we cannot use the equation ${\rho _{mass}} = \dfrac{P}{{RT}}$ .

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