Answer to Question #98275 in General Chemistry for Dakota

Question

What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuse at a rate of 3.6 mol/min

Answer 1

We set the molar masses:

gas A = 1 

gas B = 2

Then assign gas B to r₁ (and MM₁) gas A to r2 (and MM2) since we want to know the effusion rate for the gas that is twice the molar mass of the other.

Then using Graham's Law:$r1 / r2 = (MM2 / MM1) ^-1/2$

r₁ / 3.6= (1 / 2) ^-1/2

r₁ / 3.6 = 0.70710678

r₁ = 2.55 mol/min