Question

Question #fe860

Answer 1

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 `0.986` moles of liquid bromine, `"Br"_2`.

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 `"3.0128 g/cm"^3`

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

This means that every `"cm"^3` of bromine will have a mass of `3.1028` grams. Now that you know bromine's density, you need to figure out how many 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 `0.986` moles of bromine in the 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 `"100.0 kPa"` and a temperature of `0^@C"`. Under these conditions, one mole of any ideal gas occupies exactly `"22.71 L"` - this is known as the molar volume of a gas at STP.

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")`