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
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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