Question

In an experiment conducted to study the diffusion of gases using same experimental conditions, following data were recorded.

Gas A: 50 cm3 of gas A takes 7 minutes to diffuse from one container to the adjacent container.

Gas B: 50 cm3 of gas B takes 10 minutes to diffuse from one container to the adjacent container.

i. What is the rate of diffusion of gas A?

ii. What is the rate of diffusion of gas B?

iii. Which gas has higher molecular mass?

Answer 1

i. Volume of gas A diffused = 50 cm³
Time required for diffusion = 7 minutes = 7 × 60 seconds
\(Rate\ of\ diffusion\ of\ a\ gas =\frac{Volume\ of \ gas \ diffused}{Time \ required}\)

\(\frac{50 cm^3}{7\times 60 s}=0.12cm^3s^{-1}\)
∴ The rate of diffusion of gas A is 0.12 cm³ s^-1.
ii. The rate of diffusion of gas B is 0.083 cm³ s^-1.
iii. Gas B has higher molecular mass.