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Question -

From a certainapparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s-1.The diffusion of another gas under the same conditions is measured to have anaverage rate of 7.2 cm3 s-1. Identify the gas.

[Hint: Use Graham’s law ofdiffusion: R1/R2 = ( M2 /M1 )1/2, where R1 , R2 arediffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is asimple consequence of kinetic theory.]



Answer -

Given

Rate of diffusion of hydrogen, R1 = 28.7 cm3s-1

Rate of diffusion of another gas, R2 = 7.2 cm3s-1

According to Graham’s Law of diffusion,

We have,

R1 /R2 = √M2 / M1

Where,

M1 isthe molecular mass of hydrogen = 2.020g

M2 isthe molecular mass of the unknown gas

Hence,

M2 =M1 (R1 / R2)2

= 2.02 (28.7 / 7.2)2

We get,

= 32.09 g

32 g is the molecular mass of oxygen.

Therefore, the unknown gas is oxygen.

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