Question

X ml of ${{H}_{2}}$ gas effuses through a hole in a container in 5 sec. The time taken for the effusion of the same volume of the gas specified below under identical conditions is:
A. 10 sec He
B. 20 sec ${{O}_{2}}$
C. 25 sec $C{{O}_{2}}$
D. 55 sec $C{{O}_{2}}$

Answer 1

As per Graham’s law rate of diffusion (R) of gas is inversely proportional to the square root of the molar mass of the gas. It can be represented in the mathematical formula as follows.
$Rate\text{ }of\text{ }diffusion\text{ }(R)\propto \dfrac{1}{\sqrt{Molar\text{ }mass}}$

Complete step by step answer:
- In the question it is given that X ml of ${{H}_{2}}$ gas effuses through a hole in a container in 5 sec.
- Now we have to find the same volume of the gas effused through the container from the given options.
- The Graham’s law of diffusion of gas formula
$Rate\text{ }of\text{ }diffusion\text{ }(R)\propto \dfrac{1}{\sqrt{Molar\text{ }mass}}$
We know that
$R=\dfrac{volume\text{ }effused}{time\text{ }taken}=\dfrac{V}{t}$
- Therefore
$\dfrac{V}{t}\propto \dfrac{1}{\sqrt{M}}$
Here V = volume of the gas effused
t = time taken to effuse
M = Molar mass of the gas.
- In the question it is given that the volumes of the gas are same then

& t\propto \sqrt{M} \\\ & \dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ \end{aligned}$$ Where ${{t}_{1}}$ = time taken by second gas ${{t}_{2}}$ = time taken by hydrogen gas ${{M}_{1}}$ = Molar mass of second gas ${{M}_{2}}$ = Molar mass of hydrogen gas \- Now we have to cross check all the options to get the answer. \- Coming to the options, option A, 10 sec He. After substituting all the known values in the below equation we should get 10 sec as answer then option A will be correct. $$\dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}}$$ Where ${{t}_{1}}$ = time taken by Helium gas ${{t}_{2}}$ = time taken by hydrogen gas = 5 sec ${{M}_{1}}$ = Molar mass of Helium gas = 4 ${{M}_{2}}$ = Molar mass of hydrogen gas = 2 $$\begin{aligned} & \dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ & {{t}_{1}}={{t}_{2}}\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ & {{t}_{1}}=5\sqrt{\dfrac{4}{2}} \\\ & {{t}_{1}}=5\sqrt{2}\sec \\\ \end{aligned}$$ \- The obtained time is not matching with the given option A. So, option A is not correct. \- Coming to option B, 20 sec ${{O}_{2}}$ . After substituting all the known values in the below equation we should get 20 sec as answer then option B will be correct. $$\dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}}$$ Where ${{t}_{1}}$ = time taken by oxygen gas ${{t}_{2}}$ = time taken by hydrogen gas = 5 sec ${{M}_{1}}$ = Molar mass of oxygen gas = 32 ${{M}_{2}}$ = Molar mass of hydrogen gas = 2 $$\begin{aligned} & \dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ & {{t}_{1}}={{t}_{2}}\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ & {{t}_{1}}=5\sqrt{\dfrac{32}{2}} \\\ & {{t}_{1}}=5\sqrt{16} \\\ & {{t}_{1}}=20\sec \\\ \end{aligned}$$ \- The obtained time is matching with the given option B. So, option B is correct. \- Coming to option C, 25 sec $C{{O}_{2}}$ . $$\dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}}$$ Where ${{t}_{1}}$ = time taken by carbon dioxide gas ${{t}_{2}}$ = time taken by hydrogen gas = 5 sec ${{M}_{1}}$ = Molar mass of carbon dioxide gas = 44 ${{M}_{2}}$ = Molar mass of hydrogen gas = 2 $$\begin{aligned} & \dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ & {{t}_{1}}={{t}_{2}}\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ & {{t}_{1}}=5\sqrt{\dfrac{44}{2}} \\\ & {{t}_{1}}=5\sqrt{22}\sec \\\ \end{aligned}$$ \- The obtained time is not matching with the given option C. So, option C is not correct. \- Coming to option D, 55 sec $C{{O}_{2}}$ . $$\dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}}$$ Where ${{t}_{1}}$ = time taken by carbon dioxide gas ${{t}_{2}}$ = time taken by hydrogen gas = 5 sec ${{M}_{1}}$ = Molar mass of carbon dioxide gas = 44 ${{M}_{2}}$ = Molar mass of hydrogen gas = 2 $$\begin{aligned} & \dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ & {{t}_{1}}={{t}_{2}}\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}} \\\ & {{t}_{1}}=5\sqrt{\dfrac{44}{2}} \\\ & {{t}_{1}}=5\sqrt{22}\sec \\\ \end{aligned}$$ \- The obtained time is not matching with the given option D. So, option D is not correct. \- Therefore the volume of ${{H}_{2}}$ gas effuses through a hole in a container in 5 sec is equal to the volume of ${{O}_{2}}$ gas effuses through a hole in a container in 20 sec. **\- So, the correct option is B.** **Note:** We have to check all the given options by substituting them in the Graham’s formula then only we will get the equal volume of gas which can effuse through a hole in a container like hydrogen gas did in 5 sec.