Answer:
By Charles Law the volume of 1 cm^3 at standard conditions is
V = V0 * 273 / 293 = 1 cm^3 * .932 cm^3
Reducing the temperature to 0 deg C results in a volume of .932 cm^3
1 mole of gas contains 22.4 L of gas
1 L = 1000 cm^3
1000 cm^3 / L / (.932 cm^3 / container) = 1073 containers / L
So 1 mole of gas = 1073 containers * 22.4 L /container = 24,305 L
6.02E23 molecules / mole * 1 / 24305 molecules / mole = 2.48E19
number of molecules present = 2.48 * 10^19
(Easier to calculate volume of 1 cc
1 cc = 6.02E23 / (1000 * 22.4) = 2.69E19 molecules in 1 cc
Vol of 1cc = 273 / 293 = .932
2.69E19 * .932 = 2.51 * 10^19)