# An imaginary ideal gas has a density of 3 g/L at STP. What is the molar mas of this gas?

You know that you're dealing with a sample of an unknown gas that has a density of

In order to find the gas' molar mass, you need to know two things

the mass of this sample of gasthe number of moles of gas present in this sampleNow, STP conditions imply a pressure of

#"100 kPa"# and a temperature of#0^@"C"# .You will have to use the ideal as law

#color(blue)(PV = nRT)# in order to try to find a relationship between the gas' density and its molar mass.

You know that molar mass is defined as

mass per mole, so you can say that

#M_"m" = m/n implies n = m/M_"m"# Replace the number of moles in the ideal gas law equation to get

#PV = m/M_"m" * RT# You also know that density is defined as

mass per unit of volume

#rho = m/V# Notice what happens if you divide both sides of the ideal gas law equation by

#V#

#(P * color(red)(cancel(color(black)(V))))/color(red)(cancel(color(black)(V))) = m/M_"m" * (RT)/V#

#P = underbrace(m/V)_(color(green)("density")) * (RT)/M_"m" = rho * (RT)/M_"m"# Finally, isolate the molar mass on one side of the equation to get

#M_"m" = rho * (RT)/P# Now all you have to do is plug in your values - remember that

#R = 0.082("atm" * "L")/("mol" * "K")# and that the pressure must be expressed in

atmand the temperature inKelvin!

#M_"m" = 3"g"/color(red)(cancel(color(black)("L"))) * (0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#

#M_"m" = "68.085 g/mol"# SInce you only gave one sig fig for the density of the gas, the answer will be

#M_"m" = color(green)("70 g/mol")#

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*Answer 2Sign up to view the whole answerSign up with email*The molar mass of the gas is approximately 44.8 g/mol.

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*Answer from HIX Tutor**When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

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