# Question a4823

##### 1 Answer
Jun 26, 2016

$\text{45.407 L}$

#### Explanation:

The first thing to do here is use Avogadro's number to convert the number of molecules of nitrogen gas, ${\text{N}}_{2}$, to moles of nitrogen gas.

As you know, Avogadro's number is basically the definition of one mole

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"molecules} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that in order to have one mole of a molecular substance, you need to have $6.022 \cdot {10}^{23}$ molecules of that substance.

In your case, you have enough molecules of nitrogen gas to account for approximately $2$ moles, since

12.046 * 10^(23)color(red)(cancel(color(black)("molecules N"_2))) * "1 mole N"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules N"_2)))) = "2.0003 moles N"_2

Now, STP conditions are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions, one mole of any ideal gas occupies $\text{22.7 L}$ $\to$ this is known as the molar volume of a gas at STP.

Since your sample contains approximately $2$ moles of nitrogen gas, its volume will be

2.0003 color(red)(cancel(color(black)("moles N"_2))) * overbrace("22.7 L"/(1color(red)(cancel(color(black)("mole N"_2)))))^(color(purple)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)color(black)("45.407 L")color(white)(a/a)|)))#

I'll leave the answer rounded to five sig figs, the number of sig figs you have for the number of molecules of nitrogen gas.

SIDE NOTE A lot of text books and online sources still use the old definition of STP conditions, for which pressure is $\text{1 atm}$ and temperature is ${0}^{\circ} \text{C}$.

Under these conditions, one mole of any ideal gas occupies $\text{22.4 L}$.

If this is the STP definition given to you, simply redo the last calculation using $\text{22.4 L}$ instead of $\text{22.7 L}$.