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Hint: To answer this question, we must recall Graham’s law of diffusion. It was experimentally determined that the rate of diffusion of a gas is inversely proportional to the square root of the mass of their particles.

Complete step by step answer:

In the process of diffusion, two or more gases mix gradually to ultimately form a gas mixture of uniform composition. According to Graham’s Law, the ratio of the diffusion rates of two gases is the square root of the inverse ratio of their molar masses. This can be better illustrated by the formula given below:

\[\dfrac{{Rate\,of\,diffusion\,X}}{{Rate\,of\,diffusion\,Y}} = \sqrt {\dfrac{{{M_Y}}}{{{M_X}}}} \]

Since, density is directly proportional to mass and we are assuming both the gases to be of same volume, the equation takes the form:

\[\dfrac{{Rate\,of\,diffusion\,X}}{{Rate\,of\,diffusion\,Y}} = \sqrt {\dfrac{{{D_Y}}}{{{D_X}}}} \]; where D represents the density.

From the question we find that X diffuses 3 times faster than Y.

If we assume that the rate of diffusion of gas Y to be y, the rate of diffusion of gas X becomes 3y.

Substituting these values in the formula given above,

\[\dfrac{{3y}}{y} = \sqrt {\dfrac{{{D_Y}}}{{{D_X}}}} \]

On squaring both sides,

\[\dfrac{9}{1} = \dfrac{{{D_Y}}}{{{D_X}}}\]

Therefore,

\[\dfrac{{{D_X}}}{{{D_Y}}} = \dfrac{1}{9}\]

Hence, the correct answer is Option (B).

Additional information:

We should note that diffusion occurs because particles are in random motion because of their innate kinetic energies. At higher temperatures, the rate of diffusion is more since the kinetic energy of each particle is more. Hence, diffusion depends on temperature too.

Note: Graham had actually devised this law in case of effusion. It refers to the movement of gas particles along a small hole. Since, this process is similar to diffusion; Graham’s Laws are valid in this case too.

Complete step by step answer:

In the process of diffusion, two or more gases mix gradually to ultimately form a gas mixture of uniform composition. According to Graham’s Law, the ratio of the diffusion rates of two gases is the square root of the inverse ratio of their molar masses. This can be better illustrated by the formula given below:

\[\dfrac{{Rate\,of\,diffusion\,X}}{{Rate\,of\,diffusion\,Y}} = \sqrt {\dfrac{{{M_Y}}}{{{M_X}}}} \]

Since, density is directly proportional to mass and we are assuming both the gases to be of same volume, the equation takes the form:

\[\dfrac{{Rate\,of\,diffusion\,X}}{{Rate\,of\,diffusion\,Y}} = \sqrt {\dfrac{{{D_Y}}}{{{D_X}}}} \]; where D represents the density.

From the question we find that X diffuses 3 times faster than Y.

If we assume that the rate of diffusion of gas Y to be y, the rate of diffusion of gas X becomes 3y.

Substituting these values in the formula given above,

\[\dfrac{{3y}}{y} = \sqrt {\dfrac{{{D_Y}}}{{{D_X}}}} \]

On squaring both sides,

\[\dfrac{9}{1} = \dfrac{{{D_Y}}}{{{D_X}}}\]

Therefore,

\[\dfrac{{{D_X}}}{{{D_Y}}} = \dfrac{1}{9}\]

Hence, the correct answer is Option (B).

Additional information:

We should note that diffusion occurs because particles are in random motion because of their innate kinetic energies. At higher temperatures, the rate of diffusion is more since the kinetic energy of each particle is more. Hence, diffusion depends on temperature too.

Note: Graham had actually devised this law in case of effusion. It refers to the movement of gas particles along a small hole. Since, this process is similar to diffusion; Graham’s Laws are valid in this case too.

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