Question

20 L of $S{O}_2$ diffuses through a porous partition in 60 seconds. Volume of $O_2$diffuse under similar conconditionsin 30 seconds will be :

• A. 12.14 L
• B. 14.14 L
• C. 18.14 L
• D. 28.14 L

Answer 1

The correct option is B 14.14 L

The rate of diffusion is volume diffused per unit time and is inversely proportional to the square root of molar mass.

$\frac{r}{r^{\prime }}=\frac{V{t}^{\prime }}{V^{\prime }t}=\sqrt{\frac{M^{\prime }}{M}}$

Here, r, V, t, and M are the rate of diffusion, the volume diffused, the time taken for diffusion and the molar mass of $S{O}_2$
and r', V', t' and M' are the rate of diffusion, the volume diffused, the time is taken for diffusion and the molar mass of $O_2$ respectively.

Substitute values in the above equation

$\frac{20\times 30}{V^{\prime }\times 60}=\sqrt{\frac{32}{64}}$

$V^{\prime }=14.14$ L.

Hence, a volume of $O_2$ diffused under similar conditions in 30 seconds will be 14.14 L.

So, the correct option is $B$