Question

The mole fraction of oxygen in dry air near sea level is 0.20948. The concentration of

Answer 1

N sub air = ( P ) ( V ) / ( R ) ( T ) 

n sub air = ( 739 / 760 ) ( 1.000 ) / ( 0.08205 ) ( 28.5 + 273.2 ) 

n sub air = 0.03928 mol per L 

R = universal gas constant = 0.08205 atm - L per mol - deg K 

Find moles O2 per liter : 
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n sub O2 = ( y sub O2 ) ( n sub air ) 

n sub O2 = ( 0.20948 ) ( 0.03928 ) 

n sub O2 = 0.008228 moles O2 per L 

Now apply Avogadro's Number to get the O2 molecules per liter : 
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N sub O2 = ( n sub O2 ) ( N sub AVO ) 

N sub O2 = ( 0.008228 ) ( 6.022 x 10^23 ) 

N sub O2 = 4.96 x 10^21 O2 molecules per liter

It's C.

Answer 2

C,d,or e you can use the process of elimination to decide...