Question

Some moles of ${{\text{O}}_2}$ diffuse through a small opening in $18{\text{s}}$. Same number of moles of an unknown gas diffuse through the same opening in $45{\text{s}}$. Molecular mass of unknown gas is:
A. $32 \times \dfrac{{{{\left( {45} \right)}^2}}}{{{{\left( {18} \right)}^2}}}$
B. $32 \times \dfrac{{{{\left( {18} \right)}^2}}}{{{{\left( {45} \right)}^2}}}$
C. ${\left( {32} \right)^2} \times \dfrac{{45}}{{18}}$
D. ${\left( {32} \right)^2} \times \dfrac{{18}}{{45}}$

Answer 1

Hint: Graham’s law states that under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square root of their densities. Rate of diffusion is the volume of gas diffused per unit time.

Complete step by step answer:
Given data:
Time taken by ${{\text{O}}_2}$ to diffuse, ${{\text{t}}_1} = 18{\text{s}}$
Time taken by unknown gas, ${\text{t}}{ _2} = 45{\text{s}}$

Gas molecules are in a state of motion. They intermix with each other to form a homogeneous mixture. Diffusion is the random movement of molecules of gas from an area of higher concentration to the area of lower concentration. Effusion is the movement of gaseous molecules through extremely small pores in a region of low pressure.
Thomas Graham formulated the law of diffusion. This is known as Graham’s law of diffusion. It states that the velocity of a gas at a certain temperature is inversely proportional to the square root of its molecular mass.
It can be expressed as:
$\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}} = \sqrt {\dfrac{{{{\text{M}}_2}}}{{{{\text{M}}_1}}}} $, where ${{\text{r}}_1},{{\text{M}}_1}$ are the rate of diffusion and molar mass of gas $1$
${{\text{r}}_2},{{\text{M}}_2}$ are the rate of diffusion and molar mass of gas $2$.
Here gas $1$ is ${{\text{O}}_2}$ gas and gas $2$is unknown.
Rate of diffusion is the number of moles diffused divided by the time taken for diffusion.
i.e. ${{\text{r}}_1} = \dfrac{{{{\text{n}}_1}}}{{{{\text{t}}_1}}}$, where ${{\text{n}}_1}$ is the number of moles of ${{\text{O}}_2}$ and ${{\text{t}}_1}$ is the time taken.
${{\text{r}}_2} = \dfrac{{{{\text{n}}_2}}}{{{{\text{t}}_2}}}$, where ${{\text{n}}_2}$ is the number of moles of unknown gas and ${{\text{t}}_2}$ is the time taken.
It is given that time taken by ${{\text{O}}_2}$ to diffuse, ${{\text{t}}_1} = 18{\text{s}}$
Time taken by unknown gas, ${\text{t}}{ _2} = 45{\text{s}}$
Therefore the equation can be written as:
${{\text{r}}_1} = \dfrac{{{{\text{n}}_1}}}{{18{\text{s}}}}$ and ${{\text{r}}_2} = \dfrac{{{{\text{n}}_2}}}{{45{\text{s}}}}$
Therefore rate of diffusion, $\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}} = \dfrac{{{{\text{n}}_1}}}{{18}} \div \dfrac{{{{\text{n}}_2}}}{{45}}$
Taking reciprocal,
$\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}} = \dfrac{{{{\text{n}}_1}}}{{18}} \times \dfrac{{45}}{{{{\text{n}}_2}}}$
Here, ${{\text{n}}_1} = {{\text{n}}_2}$ since it mixes with each other, the ratio can be written as:
$\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}} = \dfrac{{45}}{{18}}$
Rate of diffusions is inversely proportional to the molar mass.
Therefore $\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}} = \sqrt {\dfrac{{{{\text{M}}_2}}}{{{{\text{M}}_1}}}} $
Molar mass of ${{\text{O}}_2}$ is equal to $32$. Substituting this value in the above equation, we get
$\dfrac{{45}}{{18}} = \sqrt {\dfrac{{{{\text{M}}_2}}}{{32}}} $
Squaring on both sides,
$\dfrac{{{{\left( {45} \right)}^2}}}{{{{\left( {18} \right)}^2}}} = \dfrac{{{{\text{M}}_2}}}{{32}}$
Therefore mass of unknown gas, ${{\text{M}}_2} = 32 \times \dfrac{{{{\left( {45} \right)}^2}}}{{{{\left( {18} \right)}^2}}}$

Hence, Option A is the correct option.

Additional information:
Some applications of Graham’s law are:
-When meat is sauteed with onion and garlic, the volatile substances responsible for the aroma of spices vaporize and mix with air.
-A boy is allowed to escape the air filled in a balloon, thereby it mixes with air and occupies the whole room.

Note:
Graham’s law explains that light gases effuse or diffuse very fast while heavy gases effuse or diffuse slowly. This law is derived from kinetic molecular theory. This law relates the rate of diffusion or effusion of a gas to its molar mass.