Question

Question #11d78

Answer 1

The partial pressure of ethane will be `"0.516 atm"`.

I'll base the answer on your previous question

http://socratic.org/questions/a-1-00-l-flask-is-filled-with-1-20-g-of-argon-at-25-c-a-sample-of-ethane-vapor-i#133036

There are two ways of solving for the partial pressure of ethane. The first one is to simply apply the same method used to determine the partial pressure of argon.

To put this in practice, you need to determine the number of moles of ethane present in the mixture. Since you've calculated that the total number of moles is 0.0511, and that you have 0.0300 moles of argon, the difference between these numbers will be the number of moles of ethane

`n_("ethane") = n_("total") - n_("argon") = "0.0511 - 0.0300" = "0.0211 moles"`

This means that the partial pressure of ethane will be

`P_("ethane") = chi_("ethane") * P_("total")`

`P_("ethane") = "0.0211 moles"/"0.0511 moles" * "1.25 atm" = "0.516 atm"`

The second and quicker way of finding the partial pressure of ethane is to use Dalton's law of partial pressures.

According to this, the total pressure exercited by a gas mixture that occupies a certain volume is equal to the sum of the partial pressures exercited by each gas if it alone would have occupied the same volume.

`P_("total") = P_("ethane") + P_("argon")`

`P_("ethane") = P_("total") - P_("argon") = "(1.25 - 0.734) atm" = "0.516 atm"`