A 0.532 mol sample of SO2 gas requires 52.3 s to effuse through a tiny hole. Under the same conditions, how long will it take 0.532 mol of Ar gas to effuse?

41.3 s

Explanation:

Let t₁ represent the time taken for SO₂ to effuse.

Let t₂ represent the time taken for Ar to effuse.

Let M₁ represent the molar mass of SO₂

Let M₂ represent the molar mass of Ar

From the question given above,

Time taken (t₁) for SO₂ = 52.3 s

Time taken (t₂) for Ar =?

Molar mass (M₁) of SO₂ = 32 + (16×2) = 32 + 32 = 64 g/mol

Molar mass (M₂) of Ar = 40 g/mol

Finally, we shall determine the time taken for Ar to effuse by using the Graham's law equation as shown below:

t₂ / t₁ = √(M₂ / M₁)

t₂ / 52.3 = √(40 / 64)

t₂ / 52.3 = √0.625

t₂ / 52.3 = 0.79

Cross multiply

t₂ = 52.3 × 0.79

t₂ = 41.3 s

Thus, the time taken for the amount of Ar to effuse is 41.3 s

6.4 cm (max) - 25 cm (min)