Question

A sample of oxygen gas was found to effuse at a rate equal to two times that of an unknown gas. The molecular weight of the unknown gas is ________ g/mol.

Answer 1

Answer: 128 g/mol

Step-by-step explanation:

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of the molar mass of its particles.

Mathematically, that is:

`(Rate_1)/(Rate_2)=\sqrt{(MolarMass_2)/(MolarMass_1) }`

Since, you know the ratio of two rates and the molar mass of one gas, you can calculate the molar mass of the other gas.

The molar mass of the oxygen molecule, O₂ = 2×16.0g/mol = 32.0 g/mol.

In the coming equations, I will use 32 g/mol for simplicity of writing.

`(Rate_1)/(Rate_2)=2 \\ \\ \\ MolarMass_1=32g/mol\\\\\\ 2=\sqrt{(MolarMass_2)/(32g/mol) } \\ \\ \\ 4=(MolarMass_2)/(32g/mol)\\ \\ \\ MolarMass_2=128g/mol`

So, the molecular mass of the unnknown gas is 128 g/mol.