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

Bromine is a liquid at STP, so this question is missing an assumed mass or volume. Its boiling point is #58.8^@ "C"#.

Suppose we did fill the container with exactly #"500. mL"# of bromine, then.

In that case, we use the density of bromine at #20^@ "C"# (#"3.1 g/cm"^3#) as a good approximation to the density at STP (#0^@ "C"#, #"1 bar"#) and determine the mass.

From the mass we can determine the #\mathbf("mol")#s.

#color(blue)(n_("Br"_2(l)))#

#= 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))#

#= color(blue)("9.70 mol Br"_2(l))#

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:

#PV = nRT#

#n_("Br"_2(g)) = (PV)/(RT)#

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

#= "0.0220 mol Br"_2(g)#

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.