# Question 5d930

Jun 8, 2016

${\text{63 g mol}}^{- 1}$

#### 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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, 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 $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the 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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{T \left[\text{K"] = t[""^@"C}\right] + 273.15} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Rearrange the ideal gas law equation to solve for $n$

$P V = n R T \implies n = \frac{P V}{R T}$

Plug in your values to find

$n = \left(1.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 2 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (15 + 273.15)color(red)(cancel(color(black)(} K}}}\right)$

$n = \text{0.127 moles gas}$

Now, you know that your sample of gas has a mass of $\text{8 g}$ and that it contains $0.127$ 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 $\text{63 g}$, it follows that its 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.