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Q: Chemistry - Molecular Density ( Answered,   0 Comments )
Subject: Chemistry - Molecular Density
Category: Science > Chemistry
Asked by: dodgertron-ga
List Price: $10.00
Posted: 27 Nov 2006 02:57 PST
Expires: 27 Dec 2006 02:57 PST
Question ID: 785879
how many molecules are there are there in 3 litres of OZONE, the less
common allotrope of oxygen, at standard temperature and pressure (STP,
1 atm and 298K)
Subject: Re: Chemistry - Molecular Density
Answered By: livioflores-ga on 27 Nov 2006 07:43 PST

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

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":

"Specific gas constant - Wikipedia, the free encyclopedia":

"The Individual and Universal Gas Constant":

"Forms of the Ideal Gas Law":

"Avogadro's Law":

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.


Request for Answer Clarification by dodgertron-ga on 01 Dec 2006 14:51 PST
Why is this answer different from yours.   This is the official answer
from the person that posed the question (competition entry):

8.065 X 10^22


1 mole of Ozone occupies 22.4 litres at standard temperature and pressure.
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 ÷ 22.4 x (6.022 x 1023)



Clarification of Answer by livioflores-ga on 03 Dec 2006 00:46 PST

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":

"Ideal Gas Law":

Hope this helps you to clarify this answer.

There are no comments at this time.

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