Question

The rate of diffusion of two gases A and B are in the ratio $1:4$ . A mixture contains these gases A and B in the weight ratio $2:3$ . The ratio of mole fraction of the gases A and B in the mixture is: (Assume ${P_A} = {P_B}$ )
A) $1:6$
B) $1:12$
C) $1:16$
D) $1:24$

Answer 1

Hint: We need to know that the diffusion in which movement of one substance migrates from high concentration to low concentration. We know that the rate of diffusion of gases depends on their molar mass.
$\dfrac{{{r_A}}}{{{r_B}}} = \sqrt {\dfrac{{{M_B}}}{{{M_A}}}} $
The rate of diffusion of gases A and B is represented as ${r_A}$ & ${r_B}$ respectively.
The Molar mass of gases A and B is represented as ${M_A}$ & ${M_B}$ respectively.

Complete step by step answer:
Given,
The ratio of the rate of diffusion of gases A and B is $1:4$ .
The ratio of the mass of gases A and B $\dfrac{{{m_A}}}{{{m_B}}}$ is $2:3$ .
The number of moles in A is given by,
${n_A} = \dfrac{{{m_A}}}{{{M_A}}}$
Here, ${n_A}$ is the number of moles in A.
The mass of A is ${m_A}$ and ${M_A}$ is the molar mass of A.
Similarly,
The number of moles in B is given by,
${n_B} = \dfrac{{{m_B}}}{{{M_B}}}$
Here, ${n_B}$ is the number of moles in B.
The mass of B is ${m_B}$ and ${M_B}$ is the molar mass of B.
Therefore, the ratio of mole fraction of the gases is,
$\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{{\dfrac{{{m_A}}}{{{M_A}}}}}{{\dfrac{{{m_B}}}{{{M_B}}}}}$
Now, substitute the value of the ratio of the mass of gases A and B in the above equation.
$\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{2}{3}\left( {\dfrac{{{M_B}}}{{{M_A}}}} \right)$
$\dfrac{{{M_B}}}{{{M_A}}} = \dfrac{3}{2}\left( {\dfrac{{{n_A}}}{{{n_B}}}} \right)$
Now, use the relation between the rate of diffusion of gases and the molar mass of the gases,
$\dfrac{{{r_A}}}{{{r_B}}} = \sqrt {\dfrac{{{M_B}}}{{{M_A}}}} $
$\dfrac{{{M_B}}}{{{M_A}}} = {\left( {\dfrac{{{r_A}}}{{{r_B}}}} \right)^2}$
Now, substitute the value of the ratio of the molar mass of gases A and B in the above equation.
$\dfrac{3}{2}\left( {\dfrac{{{n_A}}}{{{n_B}}}} \right) = {\left( {\dfrac{1}{4}} \right)^2}$
$\dfrac{{{n_A}}}{{{n_B}}} = {\left( {\dfrac{1}{4}} \right)^2} \times \dfrac{2}{3}$
On simplification we get,
$\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{1}{{24}}$
Therefore, the ratio of mole fraction of the gases A and B in the mixture is $1:24$ .

So, the correct answer is Option D.

Note: We must have to remember that the rate of diffusion is inversely proportional to that of the molecular weight to avoid errors while calculating the ratio between the rate diffusion of two gases. When the diffusion of gases present at constant parameters like temperature, pressure etc., we can use the Graham’s law of diffusion.