100 cm^3 of oxygen diffuses through a Porous in 3second how long will it take 150 cm^3 of sulphur (iv) oxide diffuse through the same pot? ( oxygen= 16 sulphur = 32)

3.18 seconds

Explanation:

Given the following :

Volume of oxygen (V1) = 100cm^3

Time taken (t1) = 3 seconds

Volume of Sulphur (iv) oxide (v2) = 150cm^3

From Graham's Law of diffusion:

(r1/r2) = √(m1/m2)

Where r = rate of diffusion

m = molar mass

Note rate (r) = (volume / time)

[(V1/t1) ÷ (v2/t2)] = √(m1/m2)

(v1/t1) * (t2/v2) = √(m1/m2)

m1 = 02 = (16 * 2) = 32

m2 = SO2 = (32 + (16 * 2)) = 64

(100/3) * (t2/150) = √(32/64)

100t2 / 450 = √(32/64)

100t2 / 450 = √32 / 8

100t2 / 450 = √32 / 8

100t2 * 8 = 450 * √32

800t2 = 2545.5844

t2 = 2545.5844 / 800

t2 = 3.1819805

t2 = 3.18 seconds

It will take 3.18s for 150cm^3 of Sulphur (iv) oxide to diffuse through the same pot.

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