Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm³ of SO₂ diffuses through the porous partition in 60 seconds. The volume of O₂ in dm³ which diffuses under the similar condition in 30 seconds will be: (Atomic mass of sulphur 32 u)

According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

The formula is:

r₁/r₂ = √(M₂/M₁)

Where:

r₁ = rate of diffusion of gas 1
r₂ = rate of diffusion of gas 2
M₁ = molar mass of gas 1
M₂ = molar mass of gas 2

In this case:

Gas 1 = SO₂
Gas 2 = O₂

The molar mass of SO₂ = 32 (S) + 2 * 16 (O) = 64 g/mol
The molar mass of O₂ = 2 * 16 (O) = 32 g/mol

The rate of diffusion can be expressed as volume diffused per unit time:

r₁ = V₁/t₁ = 20 dm³/60 s = 1/3 dm³/s
r₂ = V₂/t₂ = V₂/30 s

Substituting into Graham's law:

(1/3 dm³/s) / (V₂/30 s) = √(32 g/mol / 64 g/mol)

(1/3) * (30/V₂) = √(1/2)

10/V₂ = 1/√2

V₂ = 10√2 dm³

V₂ ≈ 14.1 dm³

Therefore, the volume of O₂ that diffuses in 30 seconds is approximately 14.1 dm³.