We use the old Ideal Gas equation.....
#PV=nRT#, and now all the given measurements are kosher except for the PRESSURE measurement. Pressure, #"force per unit area"# is something that is hard and non-intuitive to measure. Physical scientists, however, have long known that #"1 atmosphere"# of pressure will support a column of mercury that is #760*mm# high (and #"1 Torr"-="1 mm Hg"#). And thus we can use a measurement of length to give the pressure in atmospheres. Are you with me.....?
So here #P=(407*mm*Hg)/(760*mm*Hg*atm^-1)=0.536*atm#.
The problem is almost done now.........
#PV=nRT#; and so #n=(PV)/(RT)#
#n=(0.536*atmxx3.93*L)/(294.15*Kxx0.0821*L*atm*K^-1*mol^-1)#
#=??*atm#
Note the dimensional consistency of the answer.....
#n=(0.536*cancel(atm)xx3.93*cancelL)/(294.15*cancelKxx0.0821*cancelL*cancel(atm*K^-1)*mol^-1)#
#=1/(mol^-1)=1/(1/(mol))=mol# as required...............this is what we want, and the consistent unit is an internal check on our calculation......
