Question

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

Answer 1

`"70 g/mol"`

Explanation:

You know that you're dealing with a sample of an unknown gas that has a density of `"3 g/L"` at STP, or Standard Temperature and Pressure.

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

• the mass of this sample of gas
• the number of moles of gas present in this sample

Now, 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 atm and the temperature in Kelvin!

`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")`