Question

Calculate the relative rate of effusion of $$S{O}_2$$ to$$C{H}_4$$ under given condition:
$$(\mathrm{i})$$ under similar condition of pressure and temperature
$$(\mathrm{i}\mathrm{i})$$ through a container containing $$S{O}_2$$ and $$C{H}_4$$ in $$3:2$$ mass ratio
$$(\mathrm{i}\mathrm{i}\mathrm{i})$$ if the mixture obtained by effusing out a mixture $(n_{S{O}_2}/n_{C{H}_4}=8/1)$ for three effusing steps.
(A) $$(\mathrm{i}){\displaystyle \frac{1}{2}}(\mathrm{i}\mathrm{i}){\displaystyle \frac{3}{16}}(\mathrm{i}\mathrm{i}\mathrm{i}){\displaystyle \frac{1}{2}}$$
(B) $$(\mathrm{i}){\displaystyle \frac{1}{4}}(\mathrm{i}\mathrm{i}){\displaystyle \frac{3}{8}}(\mathrm{i}\mathrm{i}\mathrm{i}){\displaystyle \frac{1}{4}}$$
(C) $$(\mathrm{i}){\displaystyle \frac{1}{6}}(\mathrm{i}\mathrm{i}){\displaystyle \frac{5}{16}}(\mathrm{i}\mathrm{i}\mathrm{i}){\displaystyle \frac{1}{2}}$$
(D) none of these

Answer 1

Hint: As we know that, the diffusion of gases is due to the rapid movement of gas molecules and the presence of large empty space between the molecules. If the molecules are closely packed, then the rate of effusion is very less.

Complete answer
So the diffusion or effusion law is stated by Graham and it is stated as-
Under similar conditions of pressure and temperature, the rate of effusion of gases are inversely proportional to the square roots of their densities.
And we definitely know that density is directly proportional to the mass.
So, if the rate of effusion is represented as $$\mathrm{r}$$ and $$\rho$$ is the density ,$$\mathrm{n}$$ is the number of mole, $$\mathrm{x}$$ is the number of effusing steps and $$\mathrm{M}$$ is the molar mass then the rate of effusion of $$S{O}_2$$to$$C{H}_4$$ is written as-
$${\displaystyle \frac{{\mathrm{r}}_{\mathrm{S}{\mathrm{O}}_2}}{{\mathrm{r}}_{\mathrm{C}{\mathrm{H}}_4}}}=\sqrt{{\displaystyle \frac{{\rho }_{\mathrm{C}{\mathrm{H}}_4}}{{\rho }_{\mathrm{S}{\mathrm{O}}_2}}}}=\sqrt{{\displaystyle \frac{{\mathrm{M}}_{\mathrm{C}{\mathrm{H}}_4}}{{\mathrm{M}}_{\mathrm{S}{\mathrm{O}}_2}}}}={\displaystyle \frac{{{\mathrm{n}}^{\mathrm{x}}}_{\mathrm{S}{\mathrm{O}}_2}}{{{\mathrm{n}}^{\mathrm{x}}}_{\mathrm{C}{\mathrm{H}}_4}}}\sqrt{{\displaystyle \frac{{\mathrm{M}}_{\mathrm{C}{\mathrm{H}}_4}}{{\mathrm{M}}_{\mathrm{S}{\mathrm{O}}_2}}}}---(\mathrm{i})$$
The molecular mass of $$\mathrm{S}{\mathrm{O}}_2=64{\displaystyle \frac{\mathrm{g}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}}$$
The molecular mass of $$\mathrm{C}{\mathrm{H}}_4=16{\displaystyle \frac{\mathrm{g}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}}$$
The values are Putting in the above formula and we get as
$$\begin{gathered} \Rightarrow {\displaystyle \frac{{\mathrm{r}}_{\mathrm{S}{\mathrm{O}}_2}}{{\mathrm{r}}_{\mathrm{C}{\mathrm{H}}_4}}}=\sqrt{{\displaystyle \frac{16{\displaystyle \frac{\mathrm{g}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}}}{64{\displaystyle \frac{\mathrm{g}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}}}}} \\ \Rightarrow {\displaystyle \frac{{\mathrm{r}}_{\mathrm{S}{\mathrm{O}}_2}}{{\mathrm{r}}_{\mathrm{C}{\mathrm{H}}_4}}}=\sqrt{{\displaystyle \frac{1}{4}}} \\ \Rightarrow {\displaystyle \frac{{\mathrm{r}}_{\mathrm{S}{\mathrm{O}}_2}}{{\mathrm{r}}_{\mathrm{C}{\mathrm{H}}_4}}}={\displaystyle \frac{1}{2}} \end{gathered}$$
So, rate of effusion is $${\displaystyle \frac{1}{2}}$$
Let’s calculate the value of $$(\mathrm{i}\mathrm{i})$$
We are Given as mass ratio between $$\mathrm{S}{\mathrm{O}}_2$$and $$\mathrm{C}{\mathrm{H}}_4$$is $$3:2$$ means the mass of $$\mathrm{S}{\mathrm{O}}_2=3\mathrm{g}$$ and the mass of $$\mathrm{C}{\mathrm{H}}_4=2\mathrm{g}$$ so, we can calculate number of moles of $$\mathrm{S}{\mathrm{O}}_2$$and $$\mathrm{C}{\mathrm{H}}_4$$.
so, by putting these values in our $$(\mathrm{i})$$ as-
$$\begin{gathered} {\displaystyle \frac{{\mathrm{r}}_{\mathrm{S}{\mathrm{O}}_2}}{{\mathrm{r}}_{\mathrm{C}{\mathrm{H}}_4}}}={\displaystyle \frac{{\displaystyle \frac{3}{64}}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}{{\displaystyle \frac{2}{16}}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}}\sqrt{{\displaystyle \frac{16{\displaystyle \frac{\mathrm{g}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}}}{64{\displaystyle \frac{\mathrm{g}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}}}}} \\ {\displaystyle \frac{{\mathrm{r}}_{\mathrm{S}{\mathrm{O}}_2}}{{\mathrm{r}}_{\mathrm{C}{\mathrm{H}}_4}}}={\displaystyle \frac{3}{8}}\sqrt{{\displaystyle \frac{1}{4}}} \\ ={\displaystyle \frac{3}{16}} \end{gathered}$$
Let’s calculate the value of $$(\mathrm{i}\mathrm{i}\mathrm{i})$$
In this case we are Given as mole ratio between of $$\mathrm{S}{\mathrm{O}}_2$$and $$\mathrm{C}{\mathrm{H}}_4$$ $(n_{S{O}_2}/n_{C{H}_4}=8/1)$ so for three steps we have to first calculate ${{\mathrm{n}}^3}_{\mathrm{S}{\mathrm{O}}_2}/{{\mathrm{n}}^3}_{\mathrm{C}{\mathrm{H}}_4}$
Now,
$${\displaystyle \frac{{{\mathrm{n}}^3}_{\mathrm{S}{\mathrm{O}}_2}}{{{\mathrm{n}}^3}_{\mathrm{C}{\mathrm{H}}_4}}}={\displaystyle \frac{{\mathrm{n}}_{\mathrm{S}{\mathrm{O}}_2}}{{\mathrm{n}}_{\mathrm{C}{\mathrm{H}}_4}}}{\left[\sqrt{{\displaystyle \frac{{\mathrm{M}}_{\mathrm{C}{\mathrm{H}}_4}}{{\mathrm{M}}_{\mathrm{S}{\mathrm{O}}_2}}}}\right]}^3---(\mathrm{i}\mathrm{i})$$
By using equation $$(\mathrm{i}\mathrm{i})$$, we can get-
$$\begin{gathered} \Rightarrow {\displaystyle \frac{{{\mathrm{n}}^3}_{\mathrm{S}{\mathrm{O}}_2}}{{{\mathrm{n}}^3}_{\mathrm{C}{\mathrm{H}}_4}}}={\displaystyle \frac{8}{1}}{\left[\sqrt{{\displaystyle \frac{1}{4}}}\right]}^3 \\ ={\displaystyle \frac{1}{1}} \end{gathered}$$
Therefore, Putting the value of relative number of moles for three steps in $$(\mathrm{i})$$we get the answer as
$$\begin{gathered} \Rightarrow {\displaystyle \frac{{\mathrm{r}}_{\mathrm{S}{\mathrm{O}}_2}}{{\mathrm{r}}_{\mathrm{C}{\mathrm{H}}_4}}}={\displaystyle \frac{1}{1}}\sqrt{{\displaystyle \frac{1}{4}}} \\ ={\displaystyle \frac{1}{2}} \end{gathered}$$
Therefore, the relative rate of effusion for the three steps are $$(\mathrm{i}){\displaystyle \frac{1}{2}}(\mathrm{i}\mathrm{i}){\displaystyle \frac{3}{16}}(\mathrm{i}\mathrm{i}\mathrm{i}){\displaystyle \frac{1}{2}}$$ respectively.

Our correct option is option (A).

Note:
Effusion is the particular case of diffusion in which the gases contained in a vessel are allowed to escape through hole. Graham's law of diffusion is also valid for effusion.