Question

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

oxygen is __________ molecules per liter, assuming an atmospheric pressure of 739 torr and
a temperature of 29.5°C
A) 6.23
B) 0.00819
C) 4.93 × 1021
D) 3.75 × 1024
E) 5.07 × 1022

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 :
--------------------------------------...

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 :
--------------------------------------...

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.