Gas Stoichiometry and Gas Effusion and Diffusion

Question

In 25 minutes, 300 mL of sulfur dioxide diffused through a porous partition. Under the same conditions of temperature and pressure, 250 mL of an unknown gas Y diffused through the same partition in 9.11 minutes. Calculate the molecular mass of gas Y.

Answer 1

Molar mass SO₂ = 64 g / mol

Grahams law of effusion: rate gas 1 / rate gas 2 = square root (MW gas 2 / MW gas 1)

Let gas 1 be SO₂

Let gas 2 be Y

Rate SO₂ = 300 ml / 25 min = 12 ml / min

Rate Y = 250 ml / 9.11 min = 27.4 ml / min (thus the MW is < 64 g/ mol since it diffused faster than SO₂)

12 / 27.4 = √(MW Y / 64)

0.4380 = √(MW Y / 64)

0.192 = MW Y / 64

MW Y = 12.3 g / mol