Question

A vessel of volume 100ml contains 10% of oxygen and 90% of an unknown gas.The gases diffuse in 86 secs through a small hole of the vessel. If pure oxygen diffuses in 75 secs under same condition. Find the molecular weight of the gas.

Answer 1

The molecular weight of the unknown gas, calculated using Graham's law, is approximately 28.1 g/mol. This suggests that the unknown gas could potentially be nitrogen (N2), given its similar molecular weight.

Step-by-step explanation:

To find the molecular weight of the unknown gas, we can use Graham's law, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (M). As such, the ratio of the rates of effusion for two gases under identical conditions is equal to the inverse ratio of the square roots of their molar masses.

The formula derived from Graham's law is:

Rate1 / Rate2 = sqrt(M2 / M1)

Where Rate1 and Rate2 are the rates of effusion for the first and second gas, respectively, and M1 and M2 are the molecular weights of the first and second gas, respectively.

Given that oxygen (O2) with molecular weight 32 g/mol takes 75 seconds to diffuse and the mixture takes 86 seconds to diffuse, we can substitute these values into the equation:

75 / 86 = sqrt(M2 / 32)

(75 / 86)^2 = M2 / 32

M2 = (75 / 86)^2 * 32

M2 = 28.1 g/mol (approximated)

Therefore, the molecular weight of the unknown gas is approximately 28.1 g/mol, which hints that the unknown gas could be nitrogen (N2) since its molecular weight is 28 g/mol.