Problem: A gas mixture contains 1.26 g N2 and 0.86 g O2 in a 1.56-L container at 18 oC.You want to reference (Pages 216 - 222) Section 5.6 while completing this problem.Calculate the partial pressure of N2.

FREE Expert Solution

We are asked to calculate the partial pressure of O2.


Molar mass O2 = 32.00 g/mol

Molar mass N2 = 28.02 g/mol


Calculate moles each: 

mol O2 = 1.26 g O2 ×1 mol  O232.00 g O2  = 0.039 mol O2 mol N2 = 0.86 g N2 ×1 mol  N228.02 g N2 = 0.031 mol N2

Calculate mole fraction N2 :

XN2 = mol N2mol totalXN2 = mol N2mol totalXN2 = 0.031 mol0.039 mol + 0.031 molXN2 = 0.031 mol0.070 mol

XN2 = 0.443 


Given:

R = 0.08206 L-atm/mol-K

T = 18 oC +273.15 = 291.15 K

V= 1.56 L

n = 0.070 mol


Calculate total pressure: 

PV = nRTP = nRTVP = (0.070 mol)(0.08206  L-atm mol- K)(291.15 K)1.56 L

P = 1.07 atm


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Problem Details
A gas mixture contains 1.26 g N2 and 0.86 g O2 in a 1.56-L container at 18 oC.

You want to reference (Pages 216 - 222) Section 5.6 while completing this problem.

Calculate the partial pressure of N2.

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