__Question 13.12__:
From
a certain apparatus, the diffusion rate of hydrogen has an average
value of 28.7 cm

[Hint: Use Graham’s law of diffusion: R

^{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: R

_{1}/R_{2}= (M_{2}/M_{1})^{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.]

__Solution__:*R*

_{1 }= 28.7 cm

^{3}s

^{–1}

Rate of diffusion of another gas,

*R*

_{2}= 7.2 cm

^{3}s

^{–1}

According to Graham’s Law of diffusion, we have:

R

_{1}/ R

_{2}= (M

_{2}/ M

_{1})

^{1/2}

Where,

*M*

_{1}is the molecular mass of hydrogen = 2.020 g

*M*

_{2}is the molecular mass of the unknown gas

∴ M

_{2}= M

_{1}(R

_{1}/ R

_{2})

^{2}

= 2.02 (28.7 / 7.2)

^{2}

= 32.09 g

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

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