Question

From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s–1. The gas is

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

Answer 1

Answer: (D)

Oxygen

Answer 2

According to Graham’s law of diffusion,

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{\rho_{2}}{\rho_{1}}}$ or $\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

Here, $r_{1}$= diffusing rate of hydrogen $= 28.7cm^{3}s^{- 1},$

$r_{2} =$diffusion rate of unknown gas$= 7.2cm^{3}s^{- 1}$

$M_{1}$= molecular mass of hydrogen = 24

$\therefore\frac{28.7}{7.2} = \sqrt{\frac{M_{2}}{2}}$ or $M_{2} = \left( \frac{28.7}{7.2} \right)^{2} \times 2 = 32$

32 is molecular mass of oxygen.