Question

One litre of a gaseous mixture of two gases effuses in \[311\] seconds while \[2\] litres of oxygen takes \[20\min \]. The vapour density of gaseous mixture containing \[C{H_4}\]and \[{H_2}\] is:
A.4
B.4.3
C.3.4
D.2.15

Answer 1

Hint:In this question, we will use Graham’s Law of effusion. Effusion is the escaping of gas molecules through a tiny hole. The rate of diffusion and effusion of a gas depends on its molar mass. According to Graham’s Law “At constant pressure and temperature, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass.”
\[\dfrac{{Rat{e_{_1}}}}{{Rat{e_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]
\[Rat{e_1}\]-Rate of effusion for the first gas
\[Rat{e_2}\]-Rate of effusion for the second gas
\[{M_1}\]-Molar mass of first gas
\[{M_2}\]-Molar mass of second gas
Under similar conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore, rates of diffusion of different gases are inversely proportional to the square root of their mass densities.

Complete step by step answer:
We know that rate of effusion of gases is also given by:
\[Rate = \dfrac{V}{t}\]
\[V\]=Volume of gas effused
\[t\]=Time taken for effusion of gas
For gaseous mixture of \[C{H_4}\]and \[{H_2}\]:
Volume of gas effused \[{V_1} = 1L\]
Time taken for effusion of gas \[{T_1} = 311\sec \]
For oxygen gas:
Volume of gas effused \[{V_2} = 2L\]
Time taken for effusion of gas \[{T_2} = 20\min = 20 \times 60\sec \]
Therefore, we have:
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \dfrac{{\dfrac{{{V_1}}}{{{t_1}}}}}{{\dfrac{{{V_2}}}{{{t_2}}}}} = \dfrac{{{V_1}}}{{{t_1}}} \times \dfrac{{{t_2}}}{{{V_2}}} = \dfrac{1}{{311}} \times \dfrac{{20 \times 60}}{2} = \dfrac{{600}}{{311}}\]
Now according to Graham’s law of effusion:
\[\dfrac{{Rat{e_{_1}}}}{{Rat{e_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]
Molecular mass of oxygen=\[32grams\]
Substituting in the above equation:
\[\dfrac{{Rat{e_{_1}}}}{{Rat{e_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]
\[ \Rightarrow \dfrac{{600}}{{311}} = \sqrt {\dfrac{{32}}{{{M_1}}}} \]
\[ \Rightarrow \dfrac{{32}}{{{M_1}}} = \dfrac{{{{600}^2}}}{{{{311}^2}}} = 3.72\]
\[ \Rightarrow {M_1} = \dfrac{{32}}{{3.72}} = 8.6\]
Now this gives us the molecular mass of gaseous mixture to be \[8.6grams\]
As we know that:
Vapour density=\[\dfrac{M}{2}\](M=Molecular mass of the given gas)
Therefore, vapour density of gaseous mixture=\[\dfrac{{8.6}}{2} = 4.3\]
Hence, option B (4.3) is the correct answer.


Note:
Rate of the effusion is inversely proportional to the square root molecular mass and mass density. Vapour density is the half of molecular mass of given gas. Graham’s law is valid for diffusion as well as the effusion of gases.