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 $7.2\, cm^3\, s^{-1}$. The gas is

• A. Nitrogen
• B. Helium
• C. Argon
• D. Oxygen

Answer 1

Answer: (D)

According to Graham's law of diffusion, $\frac{r_{1}}{r_{2}} = \sqrt{\frac{\rho_{2}}{\rho_{1}}}$
or $\quad \frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
Here, $r_{1} =$ diffusion rate of hydrogen $= 28.7\,cm^{3} \,s^{-1}$.
$r_{2} =$ diffusion rate of unknown gas $= 7.2 \,cm^{3} \,s^{-1}$.
$M_{1} =$ molecular mass of hydrogen $= 2\, g$
$\therefore \quad \frac{28.7}{7.2} = \sqrt{\frac{M_{2}}{2}}$ or $M_{2} = \left(\frac{28.7}{7.2}\right)^{2}\times 2 \simeq 32\,g$
$32\, g$ is molecular mass of oxygen.