Ncert Exercises & Solutions

13.12 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: $\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$ , 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 the kinetic theory.]

Rate of diffusion of hydrogen,

R_{1}

= 28.7

{cm}^{3} s^{- 1}

Rate of diffusion of the unknown gas,

R_{2}

= 7.2

{cm}^{3} s^{- 1}

The molecular mass of hydrogen,

M_{1}

= 2.020 g

According to Graham’s Law of diffusion,

\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}

Then the molecular mass of the unknown gas,

M_{2} = M_{1} {(\frac{R_{1}}{(R_{2}})}^{2}

=2.02

{(\frac{28.7}{7 \cdot 2})}^{2}

= 32.09

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

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