posted by chelsey on .
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?
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)
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
Hope this helps!! :)