Hi!!
The variables to be used are:
P = pressure in atm
D = density of the gas (g/liters)
V = volume of the gas in liters
n = Number of moles of the gas
m = Mass of the gas
M = Molecular weight of gas (converted to grams/mole).
R = molar gas constant = 0.08206 atm*l/(mol*K)
Rs = Gas Constant for Ozone = R/M
T = temperature in kelvin
To solve this we can use the Ideal Gas Law using the specific gas
constant for the ozone (Rs). In this case the Ozone Gas Law is:
P = D*Rs*T ==> D = P/(Rs*T)
For the Ozone:
Rs = 0.08206 atm*l/(mol*K) / (3*16)g/mol = 0.0017095833 atm*l/g*K
Then:
D = 1 atm / (0.0017095833 atm*l/g*K * 298 K) =
= 1.96287873 g/l
Since D = Weight/Volume, then:
Weight = D*V = 1.96287873 g/l * 3 l = 5.88863619 g
Since the weight of one mole of ozone is 48 grams, there are in the 3 liters:
5.88863619g / 48(g/mol) = 0..12268 moles
According to the Avogadro´s Law, there are 6.0221415*10^23 molecules
of ozone per mole, since we have 0.12268 moles of ozone, there are:
6.0221415*10^23 * 0.12268 = 7.3879632*10^22 molecules of ozone in the
3 liters of the gas at the specified conditions.
See for references:
"Ideal gas law  Wikipedia, the free encyclopedia":
http://en.wikipedia.org/wiki/Universal_gas_law
"Specific gas constant  Wikipedia, the free encyclopedia":
http://en.wikipedia.org/wiki/Specific_gas_constant
"The Individual and Universal Gas Constant":
http://www.engineeringtoolbox.com/individualuniversalgasconstantd_588.html
"Forms of the Ideal Gas Law":
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Thermal/FormsIdealGasLaw.html
"Avogadro's Law":
http://www.chemistry.co.nz/avogadro.htm
Search strategy:
"standard temperature and pressure" volume gas
"specific gas constant" ozone
Avogadro´s number
I hope this helps you. Feel free to request for a clarification if you need it.
Regards,
liviofloresga 
Clarification of Answer by
liviofloresga
on
03 Dec 2006 00:46 PST
Hi!!
The problem with your given answer is that the standard temperature
and pressure, according to the International Union of Pure and Applied
Chemistry (IUPAC), are:
Standard temperature: 0°C = 273.15 K
Standard pressure = 1 atmosphere = 760 mmHg = 101.3 kPa (actually this
value was recently changed to 105 kPa).
At 273.15K (0°C) and 1 atm 1 mole of Ozone occupies 22.42 litres, this
is not true at 1 atm and 298K (this last condition is known as
Standard Laboratory Conditions  actually 298.15K). At this conditions
1 mole occupies more than 22.42 liters, it occupies 24.47 liters.
Clarified this point and following the reasoning of your given answer we have that:
1 mole of Ozone occupies 24.47 litres at standard 1 atm and 298K
1 mole of Ozone molecules contains 6.022 x 1023 molecules.
We need to know how many molecules are in 3 litres.
Therefore, we work it out as a percentage....
3 ÷ 24.47 x (6.022 x 10^23) = 7.383x10^22 molecules
This result is pretty similar to the one I found.
See also for references:
"Definitions of a mole":
http://www.ausetute.com.au/moledefs.html
"Ideal Gas Law":
http://www.ausetute.com.au/idealgas.html
Hope this helps you to clarify this answer.
Regards,
liviofloresga
