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Physical Chemistry

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18 Mar 2009 23:45:15 IST

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Pure oxygen diffuses thriugh an orifice for 200sec, whereas a mixture of oxygen and another gas containing 60% oxygen diffuses from the same orifice in 250 sec. Calculate the molecular weight of the unknown gas.

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vignesh the cooler version of newton

Blazing goIITian

Joined: 16 Mar 2008

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18 Mar 2009 23:53:24 IST

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rate of diffusion R is inversely proportinal to square root of molecular mass......

suppose in the second misture , 60% is O2 .. so, other 40% is unkown gas.. let its mass be M.. so,

mol. mass of the mixture = 60*32 + 40*M / 100 = ( 192 + 4M / 10 ) = X

now, R1 : R2 = _/M2 : _/M1

now, rate of diffusion is inversely proportional to time taken for diffusion..

250 : 200 = _/X : _/ 32

5:4 = _/X : _/32

25 : 16 = X: 32

from this,

X = 50 ....

put X = 50 in X = 192 + 4M / 10

so, 4M + 192 = 500

M = 77 ..

hence, molecular weight of unknown gas is 77 amu...

rtrikha

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Joined: 30 Apr 2008

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19 Mar 2009 17:31:50 IST

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According to Graham’s law of Diffusion –

Rate of diffusion R is inversely proportional to square root of Molecular mass

Let the molecular mass of Pure Oxygen is M₁ = 32

Molecular Mass of Mixture = M₂ = 60% is O₂ + 40% unknown gas. (Assume mass be M)

Mol. mass of the mixture = 60*32 + 40*M / 100 = ( 192 + 4M / 10 ) = X

R₁ / R₂ = _/ M₂ / _/ M₁

now, rate of diffusion is inversely proportional to time taken for diffusion..

250 : 200 = _/X : _/ 32

5:4 = _/X : _/32

25 : 16 = X: 32

So X = 50

Putting X = 50 in X = 192 + 4M / 10

4M + 192 = 500

M = 77

Therefore Molmass of Gas is 77 au or g

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