Question

How many oxygen molecules are in 22.4 liters of oxygen gas at 273k and 101.3kpa?

Answer 1

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 2

Answer:

$6.02\times 10^{23}$ 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

$PV=nRT$ (ideal gas )

$n=\frac{1.000 atm\times 22.4 L}{0.0821 atm L/mol K\times 273 K}=0.999 mol$

$1 mol=6.022\times 10^{23}$ molecules/atoms

Number of oxygen molecules : N

$N=n\times 6.022\times 10^{23}=0.999 mol\times 6.022\times 10^{23}$

$N=6.02\times 10^{23}$ molecules

$6.02\times 10^{23}$ oxygen molecules are in 22.4 liters of oxygen gas at 273 K and 101.3 kPa.