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

Answer:

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

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

$\frac{Rate A}{Rate B} = \sqrt{\frac{Molar mass B}{Molar mass A} }$

Therefore we have

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

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