Question

A sample of oxygen gas has a volume of 5.60 l at 27°c and 800.0 torr. how many oxygen molecules does it contain?

Answer 1

1.44x10^23 molecules of oxygen gas The ideal gas law is PV = nRT where P = pressure (800.0 Torr) V = volume (5.60 L) n = number of moles R = Ideal gas constant (62.363577 L*Torr/(K*mol) ) T = absolute temperature (27C + 273.15 = 300.15 K) Let's solve for n, the substitute the known values and solve. PV = nRT PV/RT = n (800.0 Torr*5.60 L)/(62.363577 L*Torr/(K*mol)*300.15 K) = n (4480 L*Torr)/(18718.42764 L*Torr/mol) = n 0.239336342 mol = n So we have 0.239336342 moles of oxygen molecules. To get the number of atoms, we need to multiply by avogadro's number, so: 0.239336342 * 6.0221409x10^23 = 1.44x10^23