# Question #aea36

##### 1 Answer

#### Answer:

#### 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

*universal gas constant*, usually given as

**absolute temperature** of the gas

Now, you have everything that you need in order to find **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

#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 **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.