Question

Under identical conditions of pressure and temperature,\[\;4L\;\] of gaseous mixture (\[{H_2}{\text{ }}and{\text{ }}C{H_4}\]) effuses through a hole in \[5\min \] whereas \[4L\] of a gas \[X\] of molecular mass \[36\] takes to \[10min\] to effuse through the same hole. The mole ratio of \[{H_2}:C{H_4}\] in the mixture is:
A.$1:3$
B.$1:2$
C.$2:3$
D.$1:1$

Answer 1

Hint:The rate of change of the concentration of gases or the mixture of gases from a container through a small orifice is known as rate of effusion or rate of diffusion. It is basically inversely proportional to the square root of the molar mass.

Complete step by step answer:
The term effusion means releasing or exit of something. The rate of effusion indicates the loss in concentration with respect to time. According to Graham’s law the rate of effusion or rate of diffusion is inversely proportional to the square root of its molecular weight. This means that the gases with higher molecular weights have low rate of diffusion or we can say they have low tendency to exit themselves from a hole.
So according to above discussions we can write our formulas;
$\dfrac{{dn}}{{dt}} \propto \dfrac{1}{{\sqrt {{\text{molecular weight}}} }}$
Where, $dn$ is the change in number of moles and $dt$ is the total time taken for effusion.
So we can transform this formula as;
${\text{Time taken}} \propto \sqrt {{\text{molecular weight}}} $
$ \Rightarrow \dfrac{{{{\text{T}}_{\text{1}}}}}{{{{\text{T}}_2}}} = \sqrt {\dfrac{{{m_1}}}{{{m_2}}}} $
Given:
The molecular mass of$X$ , ${m_1}$= $36$
Time taken by $X$ to effuse, ${T_1}$ = $10\min $
The molecular weight of the mixture, ${m_2}$= ?
Time taken by the mixture of gases to effuse, ${T_2}$= $5\min $
By putting all these values in the above derived formula;
$ \Rightarrow \dfrac{{10}}{5} = \sqrt {\dfrac{{36}}{{{m_2}}}} $
By solving we get;
$\sqrt {{m_2}} = 3$
$ \Rightarrow {m_2} = 9$
So the average molecular weight of the mixture is $9$ .
The formula to calculate the average molecular weight is:
$m = \dfrac{{{n_1} \times {m_1} + {n_2} \times {m_2}}}{{{n_1} + {n_2}}}$
Where,\[\;m\;\] is the average molar mass, \[{n_1}{\text{ }}and{\text{ }}{n_2}\] are the number of moles of different gases, \[{m_1}{\text{ }}and{\text{ }}{m_2}\] are the molar mass of gases. And ${m_2} = m$ .
${m_1}$ is the molar of methane i.e. $16$
${m_2}$ is the molar of hydrogen i.e. $2$
So by putting all the values in the formula we get:
$
  9 = \dfrac{{{n_1} \times 16 + {n_2} \times 2}}{{{n_1} + {n_2}}} \\
  9{n_1} + 9{n_2} = 16{n_1} + 2{n_2} \\
  7{n_1} = 7{n_2} \\
   \Rightarrow {n_1} = {n_2} \\
$
Hence the mole ratio of \[{H_2}:C{H_4}\]in the mixture is $1:1$ .
So the correct answer is (D).


Note:The rate is basically defined as the change in something with respect to time, in this reaction the moles of the gases are changing with respect to time so the rate is represented as $\dfrac{{dn}}{{dt}}$ where $dn$ represents the change in moles and $dt$ represents the change in time.