<span>438.0 mmHg is the partial pressure of oxygen
The ideal gas law is
PV = nRT
where
P = pressure
V = volume
n = number of moles
R = ideal gas constant (8.3144598 (L*kPa)/(K* mol) )
T = absolute temperature
Converting the temperature from C to K gives
43.4 + 273.15 = 316.55 K
Converting pressure from mmHg to kPa gives
675.9 mmHg * 0.133322387415 = 90.11260165 kPa
Solving for n in the equation for the ideal gas law, gives
PV = nRT
n = PV/(RT)
Substituting known values into the equation.
n = PV/(RT)
n = 90.11260165 kPa 16.2 L/(8.3144598 (L*kPa)/(K* mol) 316.55 K)
n = 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol)
n = 0.554656603 mol
So we have 0.554656603 moles of gas particles. Now to determine how much of that is oxygen.
Atomic weight oxygen = 15.999 g/mol
Atomic weight argon = 39.948 g/mol
Molar mass O2 = 2 * 15.999 = 31.998 g/mol
Now we need to figure out how many moles of O2 gas and Ar will both add up to the number of moles of gas particles and have the proper mass. So
M = number of moles of O2
0.554656603 - M = number of moles of Ar
and
M * 31.998 + (0.554656603 - M) * 39.948 = 19.3
Solve for M
M * 31.998 + (0.554656603 - M) * 39.948 = 19.3
M * 31.998 + 22.15742198 - M * 39.948 = 19.3
-M * 7.95 + 22.15742198 = 19.3
-M * 7.95 = -2.857421977
M = 0.359424148
So we now know that we have 0.359424148 moles of oxygen gas out of a total of 0.554656603 moles of gas particles. So oxygen gas is providing:
0.359424148 / 0.554656603 = 0.648012024 = 64.8012024% of the total pressure of 675.9 mmHg. So the partial pressure is
675.9 * 0.648012024 = 437.9913271 mmHg = 438.0 mmHg</span>
<span>Each mole contains Avagodro's number of atoms i.e. 6.023x10^23, so
3 moles x 6.023x10^23 atoms/mole = 18.069x10^23 atoms = 1.8x19^24 atoms </span>
