A gaseous mixture of O2 and N2 contains 30.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 365 mmHg?

Answer 1

First you have to find the moles of O2 and N2. To do this you must assume that this is a 100g sample. Therefore, 30.8g of nitrogen. Now you can find the moles of nitrogen by dividing this mass by nitrogen's molar mass:

30.8g/(14.01g/mol x 2) = 1.10moles N2

Next you can find the moles of oxygen by subracting 100 by the mass of nitrogen.

100-30.8g N2 = 69.2g O2

Now you can find the moles of oxygen:

69.2g/(16.0g/mol x 2) = 2.16moles O2

You can find the mole fraction of oxygen by using the equation :

O2=moles of component/total moles in mixture
=2.16moles/(2.16moles + 1.10moles)
=0.659

Lastly, you can find your partial pressure by multiplying the mole fraction of oxygen that you just found by the total pressure given in the question.

P1=O × Ptotal
=0.659 x 365mmHg
=240mmHg

Hope this helps!! :)