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

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35 degrees Celsius and 789 mm Hg.

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

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

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