Question

Helium is found to diffuse four times more rapidly than an unknown gas. What is the approximate molar mass of the unknown gas?

Answer 1

To solve this question, we need to know Graham's Laws of diffusion. This law gives us the relationship between the rate of diffusion and the Molecular Masses of the gases. These two terms are found to be inversely proportional to each other.

Complete Step By Step Answer:
We are given two gases He and Unknown gas ‘X’. Helium diffuses 4 times more rapidly than gas X. We can find the relationship between the velocities and Molar Masses, we can use the formula for RMS Velocity. The formula for RMS Velocity is: ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$
Where, R is the gas constant, M is the Molar mass of the gas and T is the temperature. The ratio of velocities of two gases are directly proportional to the ratio of gas diffusion ‘z’. Hence, considering the Two gases, the relationship between their velocities can be given as: $\dfrac{{{v_{rms}}(X)}}{{{v_{rms}}(He)}} = \dfrac{{{z_X}}}{{{z_{He}}}} = \sqrt {\dfrac{{{M_{He}}}}{{{M_X}}}}$
This is Graham's law of diffusion. The more the weight of the gas, the less or slowly it will diffuse. We are given that He diffuses 4 times faster than the gas X. Therefore, we can say that, $\dfrac{{{z_X}}}{{{z_{He}}}} = \dfrac{1}{4}$
The molar mass of He gas is $4.0026g/mol$ , substituting the values in the above equation we get,
$\dfrac{1}{4} = \sqrt {\dfrac{{4.0026}}{{{M_X}}}}$
Squaring both the sides, $\dfrac{1}{{16}} = \dfrac{{4.0026}}{{{M_X}}}$
${M_X} = 16 \times 4.0026 \approx 64g/mol$
Hence the molar mass of the unknown gas is found to be approximately 64 g/mol.

Note:
Graham's law states that, at constant pressure and temperature, molecules with lower molecular mass will diffuse faster than the molecules with higher molecular masses. He also related this to the rate of diffusion of gases. For two gases 1 and 2, the rate of diffusion of gases is inversely proportional to the square root of the molecular masses of the gases. It can be given gas: $\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \sqrt {\dfrac{{M.{M_2}}}{{M.{M_1}}}}$ .