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# A gas X diffuses three times faster than another gas Y. The ratio of their densities ${D_x}:{D_y}$is(A) 1/3(B) 1/9(C) 1/6(D) 1/12

Last updated date: 03rd Mar 2024
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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.

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).