Question

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 30 s for 1.0 L of O2 gas to effuse.Calculate the molar mass of the unknown gas.

Answer 1

The molar mass of the unknown gas is 392 g/mole

Step-by-step explanation:

Firstly we write out the parameters

Volume of the gas X that effuse for 105 s = 1.0 L

Volume of the O₂ gas that effuse for 30 s = 1.0 L

Molar mass of O₂ = 32 g/mole

Rate of effusion of the gas X = 1.0 L/105 s = 0.0095 L/s

Rate of effusion of O₂ gas = 1.0 L/30 s = 0.033 L/s

Graham's law of effusion states that

`(Rate A)/(Rate B) = \sqrt{(Molar mass B)/(Molar mass A) }`

Therefore we have

`(0.0095 L/s)/(0.033 L/s) =\sqrt{(32 g/mole)/(Molar mass of X) }` or Molar mass of X =
`(32)/(0.0816)` g/mole = 391.99 g/mole

The molar mass of the unknown gas is 391.99 g/mole ≈ 392 g/mole