Answer:
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.
