# Question #fe860

##### 1 Answer

Here's what I got.

#### Explanation:

I'll go out on a limb here and assume that you're interested in finding the volume occupied by **liquid** bromine,

As you know, bromine is a liquid at room temperature, which means that in order to be able to find its volume, you're going to have to use its density.

Bromine has a density of

http://www.rsc.org/periodic-table/element/35/bromine

This means that **every** **Grams** of bromine would contain that many moles.

To do that, use bromine's *molar mass*, which tells you what the mass of **one mole** of bromine is.

#0.986 color(red)(cancel(color(black)("moles Br"_2))) * "76.904 g"/(1color(red)(cancel(color(black)("mole Br"_2)))) = "75.83 g"#

This means that the sample's volume will be

#75.83 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(3.1028 color(red)(cancel(color(black)("g")))) = "24.44 cm"^3#

Expressed in *liters* and rounded to three sig figs, the answer will be

#24.44 color(red)(cancel(color(black)("cm"^3))) * "1 L"/(10^3color(red)(cancel(color(black)("cm"^3)))) = color(green)("0.244 L")#

**Alternative interpretation**

Another possibility would be to determine the volume occupied by *gaseous state*.

To do that, you'd need to know the conditions for pressure and temperature. Since no mention of that was made, I'll assume that the sample is at **STP**, Standard Temperature and Pressure.

**STP** conditions are defined as a pressure of **one mole** of any ideal gas occupies exactly

In this case, the sample will occupy a volume of

#0.986 color(red)(cancel(color(black)("moles"))) * "22.71 L"/(1color(red)(cancel(color(black)("mole")))) = color(green)("22.4 L")#