Question

From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm${}^3$ s${}^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm${}^3$ s${}^{-1}$. Identify the gas.[Hint: Use Graham's law of diffusion : $R_1/R_2=(M_2/M1{)}^{1/2}$, where $R_1,R_2$ are diffusion rates of gases 1 and 2, and $M_1$ and $M_2$ their respective molecular masses. The law is a simple consequence of kinetic theory.]

Answer 1

Rate of diffusion of hydrogen, $R_1=28.7c{m}^3{s}^{-1}$

Rate of diffusion of another gas, $R_2=7.2c{m}^3{s}^{-1}$

According to Graham’s Law of diffusion, we have:

$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$

where,

$M_1$ is the molecular mass of hydrogen $=2.020g$.

$M_2$ is the molecular mass of the unknown gas.

$\therefore M_2=M_1{\left(\frac{R_1}{R_2}\right)}^2$

$=2.02{\left(\frac{28.7}{7.2}\right)}^2=32.09g$

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.