A gas $X$ diffuses three times faster than another gas $Y$ the ratio of their densities i.e., $D_{x} : D_{y}$ is

1 / 3

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1 / 9

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1 / 6

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1 / 12

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Solution

The correct option is B $1 / 9$
Here we use Graham's law of diffusion.
Formula :
$\frac{R_{1}}{R_{2}} = \frac{\sqrt m m 2}{\sqrt m m 1}$
Let the rate of gas $Y$ be $r_{1}$ , then the rate of gas $X$ will be $3 r_{1}$
Let the density of $X$ and $Y$ be $D_{x}$ and $D_{y}$ respectively then substituting in the formula we have:
$\frac{r_{1}}{3 r_{1}} = \frac{\sqrt D_{x}}{\sqrt D_{y}}$
$(\frac{r_{1}}{3 r_{1}})^{2} = \frac{D_{x}}{D_{y}}$
$\frac{1}{9} = \frac{D_{x}}{D_{y}}$
The ratio is thus $1 : 9$

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