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

1 Answer
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^@ "C"58.8C.

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

In that case, we use the density of bromine at 20^@ "C"20C ("3.1 g/cm"^33.1 g/cm3) as a good approximation to the density at STP (0^@ "C"0C, "1 bar"1 bar) and determine the mass.

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

color(blue)(n_("Br"_2(l)))nBr2(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.