What is the molar mass of an ideal gas if a 0.622g sample of this gas occupies a volume of 300 ml @?

3 Answers

• start with these 2 equations

  PV = nRT

  n = mass / mw

  substitute and rearrange

  PV = (mass / mw) x RT

  mw = mass x RT / (PV)

  solve.

  mw = 0.622g x (0.08206 Latm/molK) x (35+273.15K) / ((789mmHg x 1atm/760mmHg) x (0.300L))

  mw = 50.5 g/mol

  *****

  Q’s?

  note the difference between how PhD chemists and PhD chemical engineers (me) do the math.

• Okay, so you need to use the ideal gas law equation:

  PV=nRT

  But first there are a few conversions you need to make so that it can be used in this equation.

  1) 300 mL = 0.3 L

  2) 789 mmHg = 1.038 atm

  3) R = 0.0821

  4) 35 C = 308 K

  Now just plug in and solve for “n”

  1.038 * 0.3 = n * 0.0821 * 308

  (1.038 * 0.3)/(.0821 * 308) = n

  0.012 = n

  So now you now how many moles are in the sample, so do a conversion to get g/mol (molar mass):

  0.622 g/0.012 mol = 52 g/mol

• Use the ideal gas law in the form:

  MM = dRT / P

  where MM is the molar mass of the gas, d is the density (g/L), T is the temperature (in Kelvin), P is the pressure (in atm) and R is the gas constant (0.08206 L atm / mol K)

  you can calculate the density of the gas using the information given:

  d = mass / volume = 0.622g / 0.300L = 2.07 g/mL

  Convert your temperature and pressure into the appropriate units for the equation:

  T = 308 K

  P = 1.04 atm

  Then plug everything into the equation:

  MM = (2.07 g/L)*(0.08206 L atm/mol K)*(308 K) / (1.04 atm)

  MM = 50.3 g/mol

  Source(s): I’m a PhD candidate in chemistry