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.