Question

A certain volume of Argon (mol. wt.=40) requires 45s to effuse through a hole at a certain pressure and temperature. The same volume of another gas of unknown volume requires 60s to pass through the same hole under the same condition of temperature and pressure. the molecular weight of the gas is
(A)53
(B)35
(C)71
(D)120

Answer 1

Hint: We can find the volume of another gas by using Graham’s law of effusion. According to this law, rate of the effusion will be inversely proportional to the molecular weight of the gaseous substance.

Complete step by step solution:
We can solve the given problem by using Graham’s law of effusion. Before moving onto the problem directly, let us first see what an effusion is. Effusion is a process in which the gas is said to leak or escape from the hole in a container. This hole is also called a pinhole. The diameter of the hole through which the molecule escapes should be smaller than the mean free path of the molecule. The escaping of gas takes place because of the pressure difference in the container and the exterior.
According to this Graham’s law of diffusion, at constant temperature and pressure, the molecules with lower molecular weight effuse at a faster rate compared to the molecules with higher molecular weight. In simple words we can say that rate of effusion is inversely proportional to the square root of the molecular weight of the gaseous atom or molecule.
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]…… (1)
In the problem the following are given:
Given,
Molecular weight of argon, \[{M_1} = 40\]
Time taken by the argon, \[{T_1} = 45s\]
Molecular weight of unknown gas, \[{M_2} = x\]
Time taken by unknown gas, \[{T_1} = 60s\]
As the rate of effusion increases time will decrease. Therefore, rate is inversely proportional to time. Hence the time will be directly proportional to the square root of the molecular weight of the gaseous atom.
Using equation (1) we can write
\[\dfrac{{Tim{e_1}}}{{Tim{e_2}}} = \sqrt {\dfrac{{{M_1}}}{{{M_2}}}} \]
\[\dfrac{{45}}{{60}} = \sqrt {\dfrac{{40}}{x}} \]
Taking the square on both sides
\[{\left( {\dfrac{{45}}{{60}}} \right)^2} = \dfrac{{40}}{x}\]
\[{\left( {\dfrac{3}{4}} \right)^2} = \dfrac{{40}}{x}\]
\[\dfrac{9}{{16}} = \dfrac{{40}}{x}\]
\[x = \dfrac{{40 \times 16}}{9}\]
\[x = 71.11 \approx 71\]
Therefore, the molecular weight of the unknown gas is 71.

Hence, the correct answer is option(C)71.

Additional information:
Difference between Effusion and diffusion.

DIFFUSION	EFFUSION
Movement of particles from higher concentration to lower concentration.	Movement of the gas particles through a small hole.
Occurs in solid, liquid and gaseous molecules	Occurs in gaseous molecules.
Rate is Slower	Rate is faster
Collision occurs	Collision is negligible.

Note: The Graham’s law of Effusion is usually used to compare the rate of effusion of different gases at constant temperature and pressure using the formula given.
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]