Question

Question #aea36

Answer 1

`"28. g mol"^(-1)`

Explanation:

Your starting point here will be the ideal gas law, which looks like this

`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 have everything that you need in order to find `n`, the number of moles of gas present in this sample under those conditions for pressure and temperature.

Since you already know the mass of the sample, you can use the number of moles it contains to find the molar mass of the gas, which is simply the mass occupied by one mole of this unknown gas.

So, rearrange the ideal gas law equation to solve for `n`

`PV = nRT implies n = (PV)/(RT)`

Plug in your values to get

`n = (2 color(red)(cancel(color(black)("atm"))) * 1 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")))) * 546color(red)(cancel(color(black)("K")))) = "0.04462 moles"`

Now, if this many moles of gas have a mass of `"1.249 g"`, it follows that one mole of gas will have a mass of

`1 color(red)(cancel(color(black)("mole"))) * "1.249 g"/(0.04462 color(red)(cancel(color(black)("moles")))) = "27.99 g"`

Since molar mass tells you the mass of one mole, it follows that your unknown gas will have a molar mass of

`"molar mass" = color(green)(|bar(ul(color(white)(a/a)"28 g mol"^(-1)color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the volume and pressure of the gas.