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 ```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)```
 ```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/individual-universal-gas-constant-d_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, livioflores-ga``` 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 Explaination: 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) Regards Dodgertron``` Clarification of Answer by livioflores-ga 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, livioflores-ga```