A given volume of methane diffuses in 20s. How long will it take the same...

Explanation

In order to explain this question, we need to understand the concept of diffusion and how it relates to the molecular weight of gases. Diffusion is the process of a substance spreading out to evenly occupy its container or environment. In the case of gases, lighter gases diffuse faster than heavier gases under the same conditions.

Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be represented as:

\[\frac{Rate_1}{Rate_2} = \sqrt{\frac{Molar\ Mass_2}{Molar\ Mass_1}}\]

Here, we are given that methane (CH₄) diffuses in 20 seconds, and we have to find the time taken for the same volume of sulfur (IV) oxide (SO₂) to diffuse under the same conditions. First, we need to calculate the molar mass of both gases.

The molar mass of methane (CH₄): C = 12, H = 1 (4 hydrogen atoms) = 4, so the molar mass of CH₄ = 12 + 4 = 16 g/mol.

The molar mass of sulfur (IV) oxide (SO₂): S = 32, O = 16 (2 oxygen atoms) = 32, so the molar mass of SO₂ = 32 + 32 = 64 g/mol.

Now, applying Graham's law of diffusion:

\[\frac{Rate_{CH_4}}{Rate_{SO_2}} = \sqrt{\frac{Molar\ Mass_{SO_2}}{Molar\ Mass_{CH_4}}}\]

Let the time taken for SO₂ to diffuse be x seconds. Since rate is inversely proportional to time, we can write:

\[\frac{20}{x} = \sqrt{\frac{64}{16}}\]

Solving for x:

\[x = 20 \times \sqrt{\frac{16}{64}} = 20 \times \frac{1}{2} = 10\]

So, it will take 10 seconds for the same volume of sulfur (IV) oxide to diffuse under the same conditions.