Question

4 litre of so2 diffuse through porous plug in 30 second .volume of CH4 in same time​

Answer 1

Answer:

8 liter is the correct answer

Answer 2

Volume of ${CH}_{4}$ diffuses through porous plug in 30 sec is 8 ltr.

Graham's law:

Graham's law states that the ratio of the rates of diffusion or effusion of two gases is equal to the square root of the inverse ratio of their molar masses.

$\frac{\text { Rate }_{1}}{\text { Rate }_{2}}=\sqrt{\frac{M_{2}}{M_{1}}},$

Given:

Volume of $SO_{2}$ is 4 ltr.

Time for diffusion of $SO_{2}$ is 30 sec.

Time for diffusion of ${CH}_{4}$ is 30 sec.

Explanation:

Solution

Rate of diffusion of  $SO_{2}=\frac{4000}{30} m L s^{-1}$

Rate of diffusion of  $\mathrm{CH}_{4}\left(r_{\mathrm{CH}_{4}}\right)=\frac{V}{30} m L s^{-1}$

Molar mass of  $\mathrm{CH}_{4}\left(\mathrm{M}_{\mathrm{CH}_{4}}\right)=16$

Molar mass of  $SO}_{2}\left(\mathrm{M}_{\mathrm{SO}_{2}}\right)=64$

According to Graham's law of emission :

$\begin{aligned}&\frac{r_{S O_{2}}}{r_{C H_{4}}}=\sqrt{\frac{M_{C H_{4}}}{M_{S O_{2}}}} \text{or}\frac{4000}{30} \times \frac{30}{V}=\sqrt{\frac{16}{64}}=\frac{1}{2} \\&4000=\frac{1}{2} \text { or } V=4000 \times 2=8000 m L\end{aligned}$

Hence, the volume of ${CH}_{4}$  diffuses through porous plug in 30 sec is 8 ltr.

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