# How many moles of Bromine are present in a 500 mL container at STP?

May 13, 2016

Bromine is a liquid at STP, so this question is missing an assumed mass or volume. Its boiling point is ${58.8}^{\circ} \text{C}$.

Suppose we did fill the container with exactly $\text{500. mL}$ of bromine, then.

In that case, we use the density of bromine at ${20}^{\circ} \text{C}$ (${\text{3.1 g/cm}}^{3}$) as a good approximation to the density at STP (${0}^{\circ} \text{C}$, $\text{1 bar}$) and determine the mass.

From the mass we can determine the $\setminus m a t h b f \left(\text{mol}\right)$s.

$\textcolor{b l u e}{{n}_{{\text{Br}}_{2} \left(l\right)}}$

= 500. cancel"mL" xx cancel("1 cm"^3)/cancel"1 mL" xx (3.1 cancel("g Br"_2))/cancel("cm"^3) xx ("1 mol Br"_2)/(2*79.904 cancel("g Br"_2))

$= \textcolor{b l u e}{{\text{9.70 mol Br}}_{2} \left(l\right)}$

Now, suppose we didn't know bromine was a liquid at STP. We would have incorrectly used the ideal gas law (assuming ideality) and gotten:

$P V = n R T$

${n}_{{\text{Br}}_{2} \left(g\right)} = \frac{P V}{R T}$

= (("1 bar")("0.5 L"))/(((0.083145 "L"cdot"bar")/("mol"cdot"K"))(273.15 "K")

$= {\text{0.0220 mol Br}}_{2} \left(g\right)$

which is much, much less than we expect, given that liquids are usually much denser than gases. Thus, the same volume should hold more of a substance if it is a liquid than if it is a gas.