Methane, CH4, diffuses in a given apparatus at the rate of 30 mL min-1 . At what rate would a gas with a molar mass of 100 diffuse under the same conditions?

Methane, CH4, diffuses in a given apparatus at the rate of 30 mL min-1 . At what rate would a gas with a molar mass of 100 diffuse under the same conditions?

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Answer

• Rate of diffusion for the gas would be 12 mL min-1
  Explanation:
  The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses
  Molar mass of methane = 12 + 4 = 16 g/mol
  This implies
  Rate of effusion of methane / Rate of effusion of unknown gas (X) = √molar mass of unknown gas / √molar mass of methane
  30 mL min-1 / X = √100 g/mol / √16 g/mol
  30 / X = 10 / 4
  30 / X = 2.5
  X = 30 / 2.5
  X = 12 mL min-1