
The rates of diffusion of two gases A and B are in the ratio 1:4. If the ratio of their masses present in the mixture is 2.3. The ratio of their mole fraction is:
Answer
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Hint: Mole fraction is a unit of concentration, defined to be equal to the number of moles of a component divided by the total number of moles of a solution. The mole fraction of all components of a solution, when added together, will be equal to 1.
Complete step by step answer:
It is given in the question that gas A has a molar mass of ${M_A}$ and gas B has a molar mass of ${M_B}$.
Now, the rate of the diffusion of the gases depends upon the molecular mass of the gases as under.
\[\dfrac{{{r_A}}}{{{r_B}}} = \sqrt {\dfrac{{{M_B}}}{{{M_A}}}} ............(1)\]
Now, we are given in the question that $\dfrac{{{r_A}}}{{{r_B}}} = \dfrac{1}{4}$ where ${r_A}{\text{ and }}{{\text{r}}_B}$ are the rate of diffusion of the respective gases.
We are also given that $\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{2}{3}$ where ${m_A}{\text{ and }}{{\text{m}}_B}$ are the masses of the gas.
On dividing the given mass by molar mass, we get the respective number of moles.
The number of moles can be denoted by n. So, we can write that
\[{n_A} = \dfrac{{{m_A}}}{{{M_A}}}{\text{ and }}{n_B} = \dfrac{{{m_B}}}{{{M_B}}}\]
Therefore, ratio of their moles can be given by
\[\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{{{m_A} \times {M_B}}}{{{m_B} \times {M_A}}}\]
We know that $\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{2}{3}$
So, $\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{{2 \times {M_B}}}{{3 \times {M_A}}}$
Thus, we can write that
\[\dfrac{{{M_B}}}{{{M_A}}} = \dfrac{{{n_A} \times 3}}{{{n_B} \times 2}}..........(2)\]
We can write from equation (1) that
\[\dfrac{{{M_B}}}{{{M_A}}} = {\left( {\dfrac{{{r_A}}}{{{r_B}}}} \right)^2}...........(3)\]
We can compare equation (2) and (3) and we get
\[\dfrac{{3{n_A}}}{{2{n_B}}} = {\left( {\dfrac{{{r_A}}}{{{r_B}}}} \right)^2}\]
Putting the available values of $\dfrac{{{r_A}}}{{{r_B}}}$, we get
\[\dfrac{{3{n_A}}}{{2{n_B}}} = {\left( {\dfrac{1}{4}} \right)^2}\]
So,
\[\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{2}{{48}} = \dfrac{1}{{24}}\]
Note: Remember that fraction of molecules is represented by mole fraction and as we know different molecules will have different masses. Therefore, the mole fraction is different from mass fraction.
Complete step by step answer:
It is given in the question that gas A has a molar mass of ${M_A}$ and gas B has a molar mass of ${M_B}$.
Now, the rate of the diffusion of the gases depends upon the molecular mass of the gases as under.
\[\dfrac{{{r_A}}}{{{r_B}}} = \sqrt {\dfrac{{{M_B}}}{{{M_A}}}} ............(1)\]
Now, we are given in the question that $\dfrac{{{r_A}}}{{{r_B}}} = \dfrac{1}{4}$ where ${r_A}{\text{ and }}{{\text{r}}_B}$ are the rate of diffusion of the respective gases.
We are also given that $\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{2}{3}$ where ${m_A}{\text{ and }}{{\text{m}}_B}$ are the masses of the gas.
On dividing the given mass by molar mass, we get the respective number of moles.
The number of moles can be denoted by n. So, we can write that
\[{n_A} = \dfrac{{{m_A}}}{{{M_A}}}{\text{ and }}{n_B} = \dfrac{{{m_B}}}{{{M_B}}}\]
Therefore, ratio of their moles can be given by
\[\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{{{m_A} \times {M_B}}}{{{m_B} \times {M_A}}}\]
We know that $\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{2}{3}$
So, $\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{{2 \times {M_B}}}{{3 \times {M_A}}}$
Thus, we can write that
\[\dfrac{{{M_B}}}{{{M_A}}} = \dfrac{{{n_A} \times 3}}{{{n_B} \times 2}}..........(2)\]
We can write from equation (1) that
\[\dfrac{{{M_B}}}{{{M_A}}} = {\left( {\dfrac{{{r_A}}}{{{r_B}}}} \right)^2}...........(3)\]
We can compare equation (2) and (3) and we get
\[\dfrac{{3{n_A}}}{{2{n_B}}} = {\left( {\dfrac{{{r_A}}}{{{r_B}}}} \right)^2}\]
Putting the available values of $\dfrac{{{r_A}}}{{{r_B}}}$, we get
\[\dfrac{{3{n_A}}}{{2{n_B}}} = {\left( {\dfrac{1}{4}} \right)^2}\]
So,
\[\dfrac{{{n_A}}}{{{n_B}}} = \dfrac{2}{{48}} = \dfrac{1}{{24}}\]
Note: Remember that fraction of molecules is represented by mole fraction and as we know different molecules will have different masses. Therefore, the mole fraction is different from mass fraction.
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