posted by Ben on .
A sample of nicotine weighing 0.3204 g was vaporized in a flask with a volume of 487 mL. This sample exrted pressure of 74.3 mm Hg at 20 degrees C. What is the molar mass of nicotine?
I know this has to do with N=P*V/R*T. and 1 atm=760 mm Hg.
Please help and show work. chemistry is not a good subject for me and i'm struggling to understand. thanks
yes, we'll use the ideal gas law formula.
PV = nRT
rearranging and noting that n = mass/molar mass,
n = PV/RT = m/MM
m/MM = PV/RT
we first convert the given pressure to atm, and temperature to Kelvin, and volume to L:
P = 74.3 mm Hg * (1 atm / 760 mm Hg) = 0.0978
T = 20 + 273.15 = 293.15 K
V = 487 mL = 0.487 L
substituting and solving for molar mass (MM),
0.3204/MM = 0.0978*(0.487)/(0.0821*293.15)
MM = 0.3204/[0.0978*(0.487)/(0.0821*293.15)]
MM = 161.9 g/mol
hope this helps~ :)