A vessel of volume 100 ml contains 10% O2 and 90% unknown gas . The gases diffuse in 86 sec through a small hole of the vessel . If pure oxygen under the same condition diffuses in 75 sec , find the molecular mass of unknown gas .
Answers
Explanation:
vessel of 100 ml contains mixture of 10% of O2 & 90% of the unknown gas.
The mixture diffuses in = 86 s
Pure O₂ diffuses in = 75 s
The molecular mass of pure O₂ = 32 g/mol
Let the rate of diffusion of the mixture be “r1” and that of pure O₂ be “r2”. Also, let the molecular mass of O₂ be “M2”, the molecular mass of the mixture be “M1” and the molecular mass of the unknown gas be “x” .
So, the molecular mass of the mixture,
M1 = [(90*x) + (32*10)] / 100 = 0.9x + 3.2
Using Graham’s Law of diffusion, we get
r1/r2 = √[M2/M1]
⇒ (1/86) / (1/75) = √[32/(0.9x + 3.2)]
⇒ (0.872)² = [32/(0.9x + 3.2)]
⇒ 0.760 (0.9x + 3.2) = 32
⇒ 0.684x + 2.432 = 32
⇒ x = 29.568 / 0.684 = 43.22 g/mol
Thus, the molecular mass of the unknown gas is 43.22 g/mol.
vessel of 100 ml contains mixture of 10% of O2 & 90% of the unknown gas.
The mixture diffuses in = 86 s
Pure O₂ diffuses in = 75 s
The molecular mass of pure O₂ = 32 g/mol
Let the rate of diffusion of the mixture be “r1” and that of pure O₂ be “r2”. Also, let the molecular mass of O₂ be “M2”, the molecular mass of the mixture be “M1” and the molecular mass of the unknown gas be “x” .
So, the molecular mass of the mixture,
M1 = [(90*x) + (32*10)] / 100 = 0.9x + 3.2
Using Graham’s Law of diffusion, we get
r1/r2 = √[M2/M1]
⇒ (1/86) / (1/75) = √[32/(0.9x + 3.2)]
⇒ (0.872)² = [32/(0.9x + 3.2)]
⇒ 0.760 (0.9x + 3.2) = 32
⇒ 0.684x + 2.432 = 32
⇒ x = 29.568 / 0.684 = 43.22 g/mol