Question

An equimolar mixture of Ne and Xe is accidentally placed in a container that has a tiny leak. After a short while, a very small proportion of the mixture has escaped. What is the mole fraction of Ne in the effusing gas?

Solution

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First, we have to calculate the effusion rate.

rNerXe\dfrac{r_{Ne}}{r_{Xe}}=Mr(Xe)Mr(Ne)\sqrt{\dfrac{Mr(Xe)}{Mr(Ne)}}

rNerXe\dfrac{r_{Ne}}{r_{Xe}}=131,2920,18\sqrt{\dfrac{131,29}{20,18}}

rNerXe\dfrac{r_{Ne}}{r_{Xe}}=2,55

In order to make it more simple, let`s say 1 mol of gas escaped.

Since Ne has 2,55 times bigger effusion rate, it effuese 2,55 times more. So if 1 mol of gas were to escape, the ratio Ne:Xe in it would be 2,55:1.

1/3,55=0,282

Moles of Ne: 0,282*2,55=0,718

Mole fraction of Ne in effusing gas:0,718/1=0,718

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