Calculate the density, g/cm3, of propane C3H8 at 107.0oC and 856 torr.?

Answer 1

First, use the gas law equation to find the number of moles in a volume of 1 ml (1 cm^3) under the specified conditions

solving for moles: PV=nRT ---> n = PV/RT

Getting things in proper units:

107C = 107 + 273K = 380K

856 torr = 856torr/760torr/atm = 1.13 atm

n = (1.13atm x 0.001)/(0.08206 Latm/molK x 380K) =

0.0000362 mole [one ml contains this # moles]

Now calculate the weight for one ml (cm^3):

0.0000362mole/ml x 44g/mol = 0.00159g/ml

Answer 2

I would use the ideal gas law: PV = nRT. First you must convert the temperature into Kelvin (K = Celsius + 273) and the pressure into atmospheres (760 torr = 1 atm). Plug the pressure in for P, the temperature in for T, plug in one mole for n, and R is the ideal gas constant: 0.0821 (L)(atm)/(mol)(K). Solve for V. This tells you the volume of one mole of propane at the given conditions. This will be in liters so you will have to convert to cm3 (1 L = 1000 cm3). Then convert one mole of propane into grams (1 mol C3H8 = 44.11 g C3H8). Divide the mass (grams) by the volume (cm3) to get your answer. I got 0.00159 g/cm3 (round to three sig figs).