Question

What is the molar mass of 31.3 g of gas exerting a pressure of 5.71 atm on the walls of a 3.04 L container at 297 K?

Answer 1

The molar mass of the gas is `43.946` grams per mole or `43.9` grams per mole using significant figures.

Explanation:

Use the ideal gas law `PV=nRT`.

`P` is the pressure in atmospheres (atm)

`V` is the volume in liters (L)

`n` is the number of moles

`R` is the universal gas constant (`.08206` `L*atm*K^(-1)*mol^(-1)`)

`T` is the temperature in kelvin

Notice that number of moles is `(grams)/(molar mass)` or `g/x` where x is the molar mass of the gas.

Plug it into the ideal gas law for `n`.

`PV=g/xRT`

Rearrange and isolate x.

`x=(gRT)/(PV)`

Plug everything in (excluding units)

`x= (31.3*.08206*297)/(5.71*3.04)`

`x= 43.946 g/(mol)`

(If you cancel out all the units, you will find that you will get `g/(mol)` which is the correct unit for molar mass.)

Answer 2

The molar mass of the gas is `"43.9 g/mol"`.

Explanation:

Use the ideal gas law equation to determine moles. Then divide the given mass by the moles to get molar mass.

`PV=nRT,`

where:

`P` is pressure, `V` is volume, `n` is moles, `R` is the gas constant, and `T` is the temperature in Kelvins.

Organize the data:

Known

`P="5.71 atm"`

`V="3.04 L"`

`R="0.0820575 L atm K"^(-1) "mol"^(-1)"`

`T="297 K"`

Unknown

`n"`

Calculate moles

Rearrange the ideal gas law equation to isolate `n`. Plug in the known values and solve.

`n=(PV)/(RT)`

`n=((5.71color(red)cancel(color(black)("atm")))xx(3.04color(red)cancel(color(black)("L"))))/((0.0820575color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1))xx(297color(red)cancel(color(black)("K"))))="0.712254 mol"` (I am keeping some guard digits to reduce rounding errors.)

Determine molar mass

Divide the given mass by the calculated moles.

`M=("31.3 g")/("0.712254 mol")="43.9 g/mol"` (rounded to three significant figures)