Question

It took 4.50 minutes for 1.0 L of helium gas to effuse through a porous barrier.

How long will it take for 0.50 L of NF3 gas to effuse under identical conditions?

Answer 1

Step 1: Explanation

Graham's law of diffusion (also known as Graham's law of effusion) states that the rate of effusion a gas is inversely proportional to the square root of its molar mass

i.e rate₁ / rate₂ = √ M₂ / M₁

Step 2: Calculation

Given,

Effusion rate of Helium(He) = rate_He = 1 L  / 4.50 min  

Effusion rate of NF₃  = rate_NF3 = 0.50 L / t where, t = time

molar mass of H_e ( M_He) = 4.003 g/mol

molar mass of NF₃  ( M_NF3) = 71 g/mol

By using above diffusion equation

rate₁ / rate₂ = √ M₂ / M₁

=> rate_He  / rate_NF3 = √ 71 g/mol / 4.003 g/mol

=> (1 L  / 4.50 min ) / ( 0.50 L / t )= 4.211495872

=> (1 L  / 4.50 min ) × ( t / 0.50 L )= 4.211495872 [ note : 0.50 L / t get reversed to cancel it ]

=>  t = 4.211495872 ( 4.50 min × 0.50 ) = 9.476 min

Hence, the 0.50 L of NH₃ gas will effuse in 9.476 min