Question

A mixture of hydrogen and oxygen in 3:1 volume ratio is allowed to diffuse through a porous partition. What should be the composition of the initial gas diffusing out the vessel?

Answer 1

Hint: Understand the concentration of hydrogen and oxygen molecules in the ratio of volume provided in the question. The rate of effusion can be estimated using Graham's law of diffusion. It will help you give a relation between the rate of effusion and molecular mass of the gas effused. The ratio of the two effusion rates will help you arrive at the correct answer.

Complete step by step answer:
We will try to understand Graham's law of diffusion and then apply the formula to find the answer.
-Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.
$r\propto \dfrac{1}{M}$

-The complete explanation of the law was given by the kinetic theory of gases a few years later. Graham's law is used in isolating isotopes of an element as they have different rates of diffusion.
-Rate of diffusion (r) =$\dfrac{volume\_diffused}{time\_taken}$
The final formula thus becomes,
$\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{\sqrt{{{M}_{2}}}}{\sqrt{{{M}_{1}}}}$
Where,
${{r}_{1}}$ is the rate of diffusion of first gas,
${{r}_{2}}$ is the rate of diffusion of second gas,
${{M}_{1}}$ is the molar mass of first gas,
${{M}_{2}}$ is the molar mass of second gas.
It is given to us that the volume ratio of hydrogen and oxygen is 3:1 and the molecular mass of oxygen is 16 g and Hydrogen is 2 g.
Substituting these values, we get,
$\dfrac{{{r}_{{{H}_{2}}}}}{{{r}_{{{O}_{2}}}}}=\dfrac{3}{1}\dfrac{\sqrt{32}}{\sqrt{2}}$ = 12
Thus, the composition of the initial gas diffusing out the vessel is 12:1.
So, the correct answer is “Option D”.

Note: The formula mentioned above is used when the diffusion of gases happens at the same conditions like temperature, pressure etc. Remember that the rate of diffusion is inversely proportional to molecular to avoid errors.