# 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...

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 mixture?

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### 1 Answer

From Dalton's law of partial pressures we know,

`P_i=x_i*P`

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,

`n_(O_2)/(n_(O_2)+n_(N_2))=0.0722`

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

`n_(O_2)/n_(N_2)=0.0722/(1-0.0722)=0.077818`

`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)`

`=(1.00*2)/(0.082*296)`

`=0.082399`

Putting the value of `n_(O_2)` ,

`n_(N_2)+0.077818*n_(N_2)=0.082399`

`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.

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