From a certain apparatus, the diffusion rate of hydrogen has an average value of $28.7 \mathrm{~cm}^{3} \mathrm{~s}$ 1. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2 \mathrm{~cm}^{3} \mathrm{~s}^{-1}$. Identify the gas.
[Hint: Use Graham’s law of diffusion: $R_{1} / R_{2}=\left(M_{2} / M_{1}\right)^{1 / 2}$, where $R_{1}, R_{2}$ are diffusion rates of gases 1 and 2 , and $\mathbf{M}_{1}$ and $\mathbf{M}_{2}$ their respective molecular masses. The law is a simple consequence of kinetic theory.]

Rate of diffusion of hydrogen is given as $R_{1}=28.7 \mathrm{~cm}^{3} \mathrm{~s}^{-1}$

Rate of diffusion of another gas is given as $R_{2}=7.2 \mathrm{~cm}^{3} \mathrm{~s}^{-1}$

According to Graham’s Law of diffusion, we can write,

$\mathrm{R}_{1} / \mathrm{R}_{2}=\sqrt{\mathrm{M}_{2}} / \mathrm{M}_{1}$

Where,

$M_{1}$ is the molecular mass of hydrogen having value $2.020 \mathrm{~g}$

$\mathrm{M}_{2}$ is the molecular mass of the unknown gas

Hence,

$\mathrm{M}_{2}=\mathrm{M}_{1}\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}$

$=2.02(28.7 / 7.2)^{2}$

We get,

$=32.09 \mathrm{~g}$

$32 \mathrm{~g}$ is the molecular mass of oxygen.

As a result, the unknown gas is oxygen.