Question

Given,\[50mL\] of hydrogen diffuses out through a small hole from a vessel in \[20minutes\] . The time needed for \[40mL\] of oxygen to diffuse out is:
(A) \[12min\]
(B) $64\min $
(C) $8\min $
(D) $32\min $

Answer 1

Hint: In order to answer this question, first we will rewrite the given facts and then we will find the rate of diffusion of both the given gases. And then we will apply the ratio between the rates of both the gases to find the required time needed for \[40mL\] of oxygen to diffuse out.

Complete answer:
The quantity of hydrogen is \[50mL\] , diffuses in \[20minutes\] .
So, the rate of diffusion of hydrogen:
$Rate({H_2}) = \dfrac{{50}}{{20}} = 2.5$
Again, the quantity of oxygen is \[40mL\] ,
So, the rate of diffusion of oxygen:
$Rate({O_2}) = \dfrac{{40}}{t}$
where, $t$ is the time taken of the diffusion of \[40mL\] of oxygen.
And we have to find the time taken to diffuse the given quantity of oxygen-
$\therefore \dfrac{{Rate({H_2})}}{{Rate({O_2})}} = \dfrac{{2.5}}{{\dfrac{{40}}{t}}} = \sqrt {\dfrac{{{M_{{O_2}}}}}{{{M_{{H_2}}}}}} = \sqrt {\dfrac{{32}}{2}} = 4$
where, ${M_{{O_2}}}$ is the molar mass of the oxygen.
${M_{{H_2}}}$ is the molar mass of the hydrogen.
Now,
$
   \Rightarrow \dfrac{{2.5}}{{\dfrac{{40}}{t}}} = 4 \\
   \Rightarrow 2.5t = 160 \\
   \Rightarrow t = 64\min \\
 $
Therefore, the time needed for \[40mL\] of oxygen to diffuse out is $64\min $ .
Hence, the correct option is (B) $64\min $ .

Note:
To comprehend Graham's law of diffusion, it is necessary to first comprehend the definition and rate of diffusion. Though the terms diffusion and effusion are frequently used interchangeably, they have very different meanings. Diffusion is the process through which particles from one gas travel to another gas.