Explain how the rate of diffusion of a gas is related to its molar mass.

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Explanation:

Graham's law of effusion (also called Graham's law of diffusion) was formulated by Scottish physical chemist Thomas Graham in 1848.[1] Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.[1] This formula can be written as:

{\displaystyle {{\mbox{Rate}}_{1} \over {\mbox{Rate}}_{2}}={\sqrt {M_{2} \over M_{1{{\mbox{Rate}}_{1} \over {\mbox{Rate}}_{2}}={\sqrt {M_{2} \over M_{1}}},

where:

Rate1 is the rate of effusion for the first gas. (volume or number of moles per unit time).

Rate2 is the rate of effusion for the second gas.

M1 is the molar mass of gas 1

M2 is the molar mass of gas 2.

Graham's law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. Thus, if the molecular weight of one gas is four times that of another, it would diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate of the other (heavier gases diffuse more slowly). A complete theoretical explanation of Graham's law was provided years later by the kinetic theory of gases. Graham's law provides a basis for separating isotopes by diffusion—a method that came to play a crucial role in the development of the atomic bomb.[2]

Graham's law is most accurate for molecular effusion which involves the movement of one gas at a time through a hole. It is only approximate for diffusion of one gas in another or in air, as these processes involve the movement of more than one gas.[2]

In the same conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore, the rates of diffusion of different gases are inversely proportional to the square roots of their mass densities.

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the answer is the first option

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