163k views
2 votes
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)​

1 Answer

1 vote

Answer:

3.18 seconds

Step-by-step 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.

User Monir
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.