Question

A certain gas takes three times as long to effuse out as helium. Its molecular mass will be

A) 36u

B) 64u

C)  9u

D) 27u

Answer 1

Hint: Graham law of effusion is applied which states that rate of any effusion of any gas is inversely proportional to the square root of its molecular mass or its density.

Complete answer:

When any gas is passed through the path which is smaller than the mean free path of the particles then the phenomena effusion occurs. 

It is not like diffusion. During effusion, collisions are the average mean path travelled. And at one time only one particle passes through effusion. 

Graham law of effusion states that effusion of any gas is inversely proportional to the square of its molecular mass or its density. In mathematical form it is written as:

 $\dfrac{{{r}_ {1}}} {{{r}_ {2}}} =\sqrt{\dfrac{{{M}_ {1}}} {{{M}_ {2}}}} $

Where, r stands for rate of diffusion of gas, M is molar mass

rate of effusion of helium is “ ${{r}_{1}}$”

rate of effusion of the unknown gas is “ ${{r}_{2}}$”

A certain gas takes three times as long to effuse out as helium, therefore the rate of effusion of helium is three times the rate of effusion of unknown gas.

${{r}_{1}}=3\times {{r}_{2}}$

${{r}_{1}}$ is directly proportional to \[\dfrac{1}{\sqrt{{{M}_{1}}}}\] where $M_1$ is the molar mass of helium.

and ${{r}_{2}}$ is directly proportional to \[\dfrac{1}{\sqrt{{{M}_{2}}}}\] where $M_2$ is molar mass of unknown gas.

Therefore,

$\dfrac{{{r}_ {1}}} {{{r}_ {2}}} =\sqrt{\dfrac{{{M}_ {2}}} {{{M}_ {1}}}} $ 

$\dfrac{3{{r}_ {1}}} {{{r}_ {1}}} =\sqrt{\dfrac{{{M}_ {2}}} {{{M}_ {1}}}} $

$\sqrt{{{M}_ {2}}} =3\sqrt{{{M}_ {1}}} $ 

Square both the sides,

${{M}_ {2}} =9{{M}_ {1}} $

Molar mass of helium is 4

${{M}_ {2}} =9\times 4=36\, g/mol$

Molar mass of an unknown is $36\, g/mol$.

The correct option is (A).

Note: Graham law of effusion is applied in such questions in which effusion of any gas is inversely proportional to its molecular mass.