Question

What is the molar mass of a gas that has a density of 1.02 g/L at 0.990 atm pressure and 37°C?

Answer 1

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

Explanation:

Your starting point here will be the ideal gas law equation

`color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "`, where

`P` - the pressure of the gas
`V` - the volume it occupies
`n` - the number of moles of gas
`R` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the gas

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas, `rho`, under those conditions for pressure and temperature, and its molar mass, `M_M`.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass `m` of this gas, you can express its molar mass as the ratio between `m` and `n`, the number of moles it contains

`color(blue)(|bar(ul(color(white)(a/a)color(black)(M_M = m/n)color(white)(a/a)|)))" " " "color(orange)((1))`

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass `m` of this gas, you can express its density as the ratio between `m` and the volume it occupies

`color(blue)(|bar(ul(color(white)(a/a)color(black)(rho = m/V)color(white)(a/a)|)))" " " "color(orange)((2))`

Plug equation `color(orange)((1))` into the ideal gas law equation to get

`PV = m/M_M * RT`

Rearrange to get

`PV * M_M = m * RT`

`P * M_M = m/V * RT`

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

Finally, use equation `color(orange)((2))` to write

`M_M = rho * (RT)/P`

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

`M_M = 1.02 "g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 37)color(red)(cancel(color(black)("K"))))/(0.990color(red)(cancel(color(black)("atm"))))`

`M_M = color(green)(|bar(ul(color(white)(a/a)color(black)("26.3 g mol"^(-1))color(white)(a/a)|)))`

I'll leave the answer rounded to three sig figs.