Chlorine is contained in a 4 liter vessel at 3 atm and 30 degrees C. How many chlorine molecules are in the vessel? (Remember that chlorine is diatomic).

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There are two steps to solving this problem:

Step 1. Use the ideal gas equation PV=nRT to find the number of moles of chlorine gas:

PV=nRT

n=PV/RT

V=4 L, P=3 atm, T=30 degrees C = 303 K, R=0.0821 L-atm/mol-K

n = (3 atm)(4 L)/(0.0821 L-atm/mol-K)(303 K) = 0.482 moles

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There are two steps to solving this problem:

Step 1. Use the ideal gas equation PV=nRT to find the number of moles of chlorine gas:

PV=nRT

n=PV/RT

V=4 L, P=3 atm, T=30 degrees C = 303 K, R=0.0821 L-atm/mol-K

n = (3 atm)(4 L)/(0.0821 L-atm/mol-K)(303 K) = 0.482 moles

Step 2. Use Avogadro's Number, 6.02x10^23, to convert moles to molecules:

(0.482 mol)(6.02x10^23 molecules/mol) = 2.90 x10^23 molecules of Cl2

A couple of notes:

  • Temperature was converted in Step 1 because it must be expressed in Kelvins for calculations using the ideal gas law.
  • Chlorine gas being diatomic isn't a factor since we're finding the number of gas particles and the particles in this case are molecules. If you needed to calculate the number of chlorine atoms present you would multiply the number of molecules by 2.

 

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