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.]
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.