# Question #5d930

##### 1 Answer

#### Explanation:

Your strategy here will be to use the **ideal gas law** equation to find the **number of moles** of gas present in that sample.

Once you know that, you can use the given mass of gas to determine the mass of *a single mole*, which in what the **molar mass** of the gas essentially is.

So, the ideal gas law equation 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

The first thing to notice here is that you need to have the temperature expressed in *Kelvin*, **not** in *degrees Celsius*, so make sure that you convert that by using

#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#

Rearrange the ideal gas law equation to solve for

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

Plug in your values to find

#n = (1.5 color(red)(cancel(color(black)("atm"))) * 2 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")))) * (15 + 273.15)color(red)(cancel(color(black)("K))))#

#n = "0.127 moles gas"#

Now, you know that your sample of gas has a mass of **moles** of gas. In order to find the *molar mass* of the gas, you must find the mass of **one mole** by using

#1 color(red)(cancel(color(black)("mole gas"))) * "8 g"/(0.127color(red)(cancel(color(black)("moles gas")))) = "63 g"#

So, if **one mole** of gas has a mass of *molar mass* is equal to

#"molar mass" = color(green)(|bar(ul(color(white)(a/a)color(black)("63 g mol"^(-1))color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**.