If helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is:

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Solution

The correct option is C 2
Helium atomic weight $= 4 g$ .
Methane molecular weight $C H_{4} = 12 + 4 (1) = 16 g$ .
$Rate of diffusion \propto \frac{1}{\sqrt{molecular weight}}$
$\frac{(Rate of diffusion) C H_{4}}{(Rate of diffusion) H e} = \sqrt{\frac{4}{16}} = \frac{1}{2}$
$∴$ Rate of diffusion of Helium is $2$ times that of $C H_{4}$ .

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If helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is:

The correct option is C 2Helium atomic weight =4g.Methane molecular weight CH4=12+4(1)=16g.Rate of diffusion∝1√molecular weight(Rate of diffusion) CH4(Rate of diffusion)He=√416=12∴ Rate of diffusion of Helium is 2 times that of CH4.