Question

At ${1200^ \circ }{{C}}$, mixture of ${{C}}{{{l}}_2}$ and ${{Cl}}$ atoms (both in gaseous state) effuses $1.16$ times faster than Krypton effuses under identical conditions. The fraction of chlorine molecules dissociated into atoms is approximate:
[Molecular weight of ${{Kr = 84}}$]
A. $13\% $
B. $22\% $
C. $87\% $
D. $44\% $

Answer 1

Hint: Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in a container. Graham’s law of effusion relates the rate of effusion of a gas with the density of the gas.

Complete step by step answer:
Gas molecules are in a state of constant motion. So they intermix with each other to form a homogeneous mixture. Effusion is a process in which a gas under pressure escape out a fine hole. Graham’s law of effusion 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.
Mathematically it can be represented as:
${{r}}\alpha \dfrac{1}{{\sqrt {{d}} }}$, where ${{r}}$ is the rate of diffusion and ${{d}}$ is the density of gas.
It can also be expressed as $\dfrac{{{{{r}}_1}}}{{{{{r}}_2}}} = \sqrt {\dfrac{{{{{d}}_2}}}{{{{{d}}_1}}}} $
Molecular weight, ${{M}}$ is twice the density value.
Thus, \[\dfrac{{{{{r}}_1}}}{{{{{r}}_2}}} = \sqrt {\dfrac{{\dfrac{{{{{M}}_2}}}{2}}}{{\dfrac{{{{{M}}_1}}}{2}}}} = \sqrt {\dfrac{{{{{M}}_2}}}{{{{{M}}_1}}}} \]
The rate of diffusion is defined as the volume of gas diffused per unit time.
It is given that the ratio of \[\dfrac{{{{{r}}_1}}}{{{{{r}}_2}}}\] is $1.16$, where ${{{r}}_1}$ is the rate of effusion of mixture of ${{C}}{{{l}}_2}$ and ${{Cl}}$ atoms and ${{{r}}_2}$ is the rate of effusion of Krypton.
The molecular weight of Krypton, ${{{M}}_2} = 84$
Molecular weight of the mixture of ${{C}}{{{l}}_2}$ and ${{Cl}}$ atoms, ${{{M}}_1}$ can be calculated from the formula given below:
\[1.16 = \sqrt {\dfrac{{84}}{{{{{M}}_1}}}} \]
On squaring on both sides, we get
\[{\left( {1.16} \right)^2} = \dfrac{{84}}{{{{{M}}_1}}}\]
On simplifying, we get
$1.35 = \dfrac{{84}}{{{{{M}}_1}}} \Leftrightarrow {{{M}}_1} = \dfrac{{84}}{{1.35}} = 62.2$
Since ${{{M}}_1}$ is the average molecular weight of ${{C}}{{{l}}_2}$ and mixture.
We have ${{C}}{{{l}}_2} \rightleftharpoons 2{{C}}{{{l}}^ - }$
Initial $1 \;\;\;\;\;\;\;\;\;\;\; \;\;\; 0$
Final $1 - {{x}} \;\;\;\;\;\;\; 2{{x}}$
Total number of moles, ${{{n}}_{{{tot}}}} = 1 - {{x}} + 2{{x = 1 + x}}$
Hence, $
62.2 = \dfrac{{2{{x}} \times 35.5 + 71\left( {1 - {{x}}} \right)}}{{1 + {{x}}}} $
$62.2 = \dfrac{{71{{x}} + 71\left( {1 - {{x}}} \right)}}{{1 + {{x}}}} $
On simplification, we get
$62.2 = \dfrac{{71{{x}} + 71 - 71{{x}}}}{{1 + {{x}}}} = \dfrac{{71}}{{1 + {{x}}}}$
On solving by cross-multiplying, we get
$62.2 + 62.2{{x}} = 71 \Leftrightarrow 62.2{{x}} = 8.8 \Leftrightarrow {{x}} = 0.14$
i.e. $14\% $ dissociated. $13\% $ is the closest number in the given options.

So, the correct answer is Option A.

Note: Graham’s law of effusion helps in the separation of gases having different densities. Isotopes of certain elements can also be separated. Carbon dioxide is heavier than oxygen and nitrogen gases present in the air, but it does not form the lower layer of the atmosphere. The property of diffusion is independent of the force of gravitation.