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

NetherCraft 0

35 degrees Celsius and 789 mm Hg.

Question is part of my chemistry review for my exam tomorrow. Any help?

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

Also Check This  Is your browser running http version 1.0 or 1.1? what version of http is the server running?

Leave a Reply

Your email address will not be published. Required fields are marked *