Question

The rate of effusion of Xe gas through a porous barrier is observed to be

9.14 x 10-4 mol/h. Under the same conditions, the rate of effusion of Ar gas would be:
______ mol/h.​

Answer 1

Rate of Ar= 1.65 x
`10^(-3)` mol/h

Step-by-step explanation:

Here we'll use Graham's Law where
`(R1)/(R2) = \sqrt{(M2)/(M1) }`

R= rate of effusion and M= molar mass

Let's plug in what we know.

`(R1)/(9.14 x 10x^(-4) ) = \sqrt{(131 g Xe)/(40g Ar) }`

You could switch them where Xe is M1 (on the bottom) instead of M2, but I find it easier when the unknown is R1, where it acts like a whole number. It makes the algebra part of the equation easier.

Let's solve for R1.

`(R1)/(9.14 x 10x^(-4) ) = √(3.27)\\`

`(R1)/(9.14 x 10x^(-4) ) = 1.089`

R1= 1.089 x
`9.14 x 10^(-4)`

R1= 1.65 x
`10^(-3)` mol/h