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**Hint:**Gaseous particles tend to undergo diffusion because they have kinetic energy. Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy. The rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Gas with the lowest molar mass will have the highest rate of diffusion.

**Formula used:**\[r \propto \dfrac{1}{{\sqrt M }}\]

**Complete step by step answer:**

Graham’s Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.

The value of compression of two gases, the rate of diffusion through a fine hole is given by:

\[r \propto \dfrac{1}{t} \propto \dfrac{1}{{\sqrt M }}\] where r is the rate of diffusion, t is the time of diffusion and M is the molar mass.

The rate of diffusion depends on several factors such as the concentration gradient ( the increase or decrease in concentration from one point to another), the amount of surface area available for diffusion, and the distance the gas particles must travel.

Now, let the rate of diffusion of oxygen is \[{r_O} \propto \dfrac{1}{{{t_o}}} \propto \dfrac{1}{{\sqrt {{M_0}} }}\] and the rate of diffusion of the X gas is \[{r_X} \propto \dfrac{1}{{{t_X}}} \propto \dfrac{1}{{\sqrt {{M_X}} }}\]

Therefore, comparing both formulae,

\[

\dfrac{{{t_0}}}{{{t_x}}} = \dfrac{{\sqrt {{M_0}} }}{{\sqrt {{M_X}} }} \\

\dfrac{{47}}{{74}} = \dfrac{{\sqrt {32} }}{{\sqrt {{M_X}} }} \\

{M_X} = {\left( {\dfrac{{74}}{{47}}} \right)^2} \times 32 \\

{M_X} = 79.32 \\

\]

Therefore, the molar mass of the gas is \[79.32gm/mol\].

**Hence option C is correct.**

**Note:**

Diffusion is inversely proportional to molecular weight. Therefore, \[N{H_3}\] has the highest diffusion rate, whereas, carbon dioxide has the greatest molecular mass and therefore should be expected to diffuse the slowest.

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