If the rate of diffusion of oxygen gas is taken as 1. What will be
If the rate of diffusion of oxygen gas is taken as 1. What will be the rate of diffusion of methane whose relative molar mass is 16?
- A) 2.0
- B) 1.8
- C) 1.40
- D) 1.0
Correct Answer: D) 1.0
Explanation
First, let's understand diffusion. Diffusion is the process by which molecules move from an area of higher concentration to an area of lower concentration, eventually leading to equal concentrations throughout the space. In gases, diffusion is mainly influenced by the molecular weight or molar mass of the gas.
In this question, we are given the rate of diffusion of oxygen gas as 1 and asked to find the rate of diffusion of methane, whose relative molar mass is 16. To solve this, we will use Graham's law of diffusion, which states:
\[\frac{Rate_1}{Rate_2} = \sqrt{\frac{Molar\,mass_2}{Molar\,mass_1}}\]
Where Rate_1 and Rate_2 are the rates of diffusion of gas 1 and gas 2, and Molar mass_1 and Molar mass_2 are the molar masses of gas 1 and gas 2, respectively.
First, let's find the molar mass of oxygen gas. Oxygen has a molar mass of 16 g/mol, and since oxygen gas is diatomic (O2), its molar mass is 2 * 16 = 32 g/mol. Now we can apply Graham's law to find the rate of diffusion of methane (CH4):
\[\frac{Rate_{O_2}}{Rate_{CH_4}} = \sqrt{\frac{Molar\,mass_{CH_4}}{Molar\,mass_{O_2}}}\]Plugging in the given values:
\[\frac{1}{Rate_{CH_4}} = \sqrt{\frac{16}{32}}\]Now, we can solve for the rate of diffusion of methane:
\[Rate_{CH_4} = \frac{1}{\sqrt{\frac{1}{2}}}\]After simplifying, we find that the rate of diffusion of methane is equal to 1.0, which is the correct answer (Option D).
In summary, the rate of diffusion of methane is the same as the rate of diffusion of oxygen gas, which is 1.0, according to Graham's law of diffusion.