# What is the volume occupied by 3.0*10^23 molecules of bromine gas at STP?

Jul 19, 2016

$\textcolor{m a \ge n t a}{\text{11 L}}$

#### Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Because we are at STP, we will have to use the ideal gas law:}}$

At standard temperature and pressure, the temperature is 273K and the pressure is 1 atm.

Next, list your known and unknown variables. Our only unknown is the volume of $B {r}_{2} \left(g\right)$. Our known variables are P,R, and T.

We don't necessarily have $n$ yet, but we're given a value that will lead us to the number of moles of bromine gas. All we have to do is convert molecules to moles using the conversion factor below:

We'll use the second conversion factor because we can cancel out molecules and end up with moles:

3.0xx10^(23)cancel"molecules"xx(1mol)/(6.02x10^(23)cancel"molecules") $=$ $0.498 m o l$

Now all of the variables have good units! All that's left to do is rearrange the equation and solve for V like so:

$V = \frac{n \times R \times T}{P}$
$V = \left(0.498 \cancel{\text{mol"xx0.0821(Lxxcancel"atm")/(cancel"mol"xxcancel"K")xx(273cancelK))/(1cancel"atm}}\right)$
$V = 11 L$