A mixture of `N_2 ` and `O_2`  gases are in a 2.0L container at 23 degrees C and at a total pressure of 1.00 atm. The partial pressure of O2 was .0722 atm. How many grams of `N_2 ` were in the gas...

1 Answer | Add Yours

llltkl's profile pic

Posted on

From Dalton's law of partial pressures we know,


Where P is the total pressure, `P_i` is the partial pressure of the ith component, and `x_i` , its mole-fraction.

Here `P_(O_2)=x_(O_2)*P`

`rArr 0.0722=x_(O_2)*1.00`

`rArr x_(O_2)=0.0722`

Applying the concept of mole-fraction for a binary gas mixture,


`rArr n_(O_2)(1-0.0722)=0.0722n_(N_2)`


`rArr n_(O_2)=0.077818*n_(N_2)`

Agan, assuming ideal behaviour, total number of moles of gaseous species `(n_(O_2)+n_(N_2))`

`= (PV)/(RT)`



Putting the value of `n_(O_2)` ,


`rArr n_(N_2)=0.082399/(1+0.077818)=0.07645`

Mass of `N_2` = number of moles of `N_2` * molar mass of `N_2`

`=0.07645*28=2.14` g.

Therefore, mass of `N_2` present in the gas mixture was 2.14 g.


We’ve answered 327,683 questions. We can answer yours, too.

Ask a question