From a certain apparatus, the diffusion rate of hydrogen has an average value of \( 28.7 cm^3 s^{-1}\). The diffusion of another gas under the same conditions is measured to have an average rate of \( 7.2 cm^3 s^{-1}\). Identify the gas. [Hint: Use Graham’s law of diffusion: \( \dfrac{R_1}{R_2}\) = \( (\dfrac{M_2}{M_1})^ \dfrac{1}{2}\), where \( R_1 , R_2\) are diffusion rates of gases 1 and 2, and \( M_1\) and \( M_2\) their respective molecular masses. The law is a simple consequence of kinetic theory.]

Asked by Pragya Singh | 1 year ago |  163

1 Answer

Solution :-


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,

\( \dfrac{R_1}{R_2}=\sqrt{\dfrac{M_2}{M_1}}\)


M1 is the molecular mass of hydrogen = 2.020g

M2 is the molecular mass of the unknown gas


M2 = M1\(( \dfrac{R_1}{R_2})^2\)

= 2.02\( ( \dfrac{28.7}{7.2})^2\)

We get,

= 32.09 g

32 g is the molecular mass of oxygen.

Therefore, the unknown gas is oxygen.

Answered by Pragya Singh | 1 year ago

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