How many oxygen molecules are in 22.4 liters of oxygen gas at 273k and 101.3kpa
First solve the number of moles of the oxygen gas by using the ideal gas equation:
PV = nRT
Where n is the number of moles
n = PV/RT
n = (101 300 Pa) (22.4 L) (1 m3/1000 L ) / ( 8.314 Pa m3 / mol K) ( 273 K)
n = 1 mol O2
the number of molecules can be solve using avogrados number 6.022x10^23 molecule / mole
molecules of one mole O2 = 6.022x 10^23 molecules
Answer:
[tex]6.02\times 10^{23} [/tex] oxygen molecules are in 22.4 liters of oxygen gas at 273 K and 101.3 kPa.
Explanation:
Pressure of the oxygen gas = P = 101.3 kPa = 1.000 atm
1 atm = 101.3 kPa
Volume of oxygen gas = V = 22.4 L
Temperature of the oxygen gas = T = 273 K
Moles of oxygen gas = n
[tex]PV=nRT[/tex] (ideal gas )
[tex]n=\frac{1.000 atm\times 22.4 L}{0.0821 atm L/mol K\times 273 K}=0.999 mol[/tex]
[tex]1 mol=6.022\times 10^{23} [/tex] molecules/atoms
Number of oxygen molecules : N
[tex] N=n\times 6.022\times 10^{23}=0.999 mol\times 6.022\times 10^{23}[/tex]
[tex]N=6.02\times 10^{23} [/tex] molecules
[tex]6.02\times 10^{23} [/tex] oxygen molecules are in 22.4 liters of oxygen gas at 273 K and 101.3 kPa.
