Answer:
Rate of Ar= 1.65 x [tex]10^{-3}[/tex] mol/h
Explanation:
Here we'll use Graham's Law where [tex]\frac{R1}{R2} = \sqrt{\frac{M2}{M1} }[/tex]
R= rate of effusion and M= molar mass
Let's plug in what we know.
[tex]\frac{R1}{9.14 x 10x^{-4} } = \sqrt{\frac{131 g Xe}{40g Ar} }[/tex]
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.
[tex]\frac{R1}{9.14 x 10x^{-4} } = \sqrt{3.27}\\[/tex]
[tex]\frac{R1}{9.14 x 10x^{-4} } = 1.089[/tex]
R1= 1.089 x [tex]9.14 x 10^{-4}[/tex]
R1= 1.65 x [tex]10^{-3}[/tex] mol/h
