Answer to Question #336681 in General Chemistry for Oswald

Question

Gas X has a molar mass of 100 g/mol and Gas Y has a molar

mass of 32 g/mol. How much faster or slower does Gas Y

effuse from a small opening than Gas X if they are at the same

temperature?

Answer 1

Graham's Law can be expressed as:

r_X(MM_X)^1/2 = r_Y(MM_Y)^1/2

where

r_X = rate of effusion/diffusion of Gas X

MM_X = molar mass of Gas X

r_Y = rate of effusion/diffusion of Gas Y

MM_Y = molar mass of Gas Y

We want to know how much faster or slower Gas Y effuses compared to Gas X. To get this value, we need the ratio of the rates of Gas Y to Gas X. Solve the equation for r_Y/r_X.

r_Y/r_X = (MM_X)^1/2/(MM_Y)^1/2

r_Y/r_X = [(MM_X)/(MM_Y)]^1/2

Use the given values for molar masses and plug them into the equation:

r_Y/r_X = [(100 g/mol)/(32)]^1/2

r_Y/r_X = [3.125]^1/2

r_Y/r_X = 1.8

Gas Y will effuse six times faster than the heavier Gas X.