Question

Question #4b1ce

Answer 1

`"32.47 g mol"^(-1)`

Explanation:

As you know, the density of a substance, `rho`, tells you the number of grams present in one unit of volume of said substance.

In your case, you know that an unknown gas has a density of `"1.429 g L"^(-1)` at STP conditions, which are defined as

• `"pressure" = "100 kPa" = 100/101.325` `"atm"`

• `"temperature" = 0^@"C" = "273.15 K"`

This tells you that every `"1 L"` of this gas has a mass of `"1.429 g"`. So, you can say that

`color(blue)(rho = m/V)`

where `m` represents the mass of the gas and `V` the volume it occupies, tells you the density of the gas.

At this point, your tool of choice will be the ideal gas law equation

`color(blue)(ul(color(black)(PV = nRT)))`

Here

• `P` is the pressure of the gas
• `V` is the volume it occupies
• `n` is the number of moles of gas present in the sample
• `R` is the universal gas constant, equal to `0.0821("atm L")/("mol K")`
• `T` is the absolute temperature of the gas

If you use the fact that the number of moles of gas can be expressed as the ratio between the mass of the sample and the molar mass of the gas `M_M`

`n = m/M_M`

you can rewrite the ideal gas law equation as

`PV = m/M_M * RT`

Rearrange this to isolate `M_M` on one side of the equation

`M_M = color(blue)(m/V) * (RT)/P`

This is equivalent to

`M_M = color(blue)(rho) * (RT)/P`

Plug in the density of the gas and the aforementioned conditions for pressure and temperature to find the molar mass of the gas

`M_M = "1.429 g" color(red)(cancel(color(black)("L"^(-1)))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 273.15 color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))`

`color(darkgreen)(ul(color(black)(M_M = "32.47 g mol"^(-1))))`

The answer is rounded to four sig figs, the number of sig figs you have for the density of the gas.

SIDE NOTE A lot of sources still use the old definition of STP conditions

• `"pressure = 1 atm"`
• `"temperature" = 0^@"C" = "273.15 K"`

so if this is the definition given to you, redo the calculations using `"1 atm"` instead of `100/101.325` `"atm"` for the pressure of the gas.