# Question d36b2

Feb 22, 2017

$\text{15.2 mL}$

#### Explanation:

As you know, the density of a substance tells you the mass of exactly one unit of volume of that substance.

In your case, liquid bromine is said to have a density of ${\text{3.102 g mL}}^{- 1}$, which means that every $\text{1 mL}$ of bromine has a mass of $\text{3.102 g}$.

color(blue)(color(purple)("3.102 g") color(white)(.)"mL"^(-1)) -> color(purple)("3.102 g")color(white)(.) color(black)("for every") color(white)(.)color(blue)("1 mL")color(white)(.)"of liquid Br"_2

Now, your goal here is to figure out the volume of liquid bromine that would contain $0.295$ moles of bromine. In order to be able to do that, you must convert the number of moles of bromine to grams.

Bromine has a molar mass of ${\text{159.808 g mol}}^{- 1}$, which means that every $1$ mole of bromine has a mass of $\text{159.808 g}$.

In your case, the sample of liquid bromine will have a mass of

0.295 color(red)(cancel(color(black)("moles Br"_2))) * "159.808 g"/(1color(red)(cancel(color(black)("moles Br"_2)))) = "47.14 g"#

You can now use the density of liquid bromine as a conversion factor to determine exactly how many milliliters of bromine would have a mass of $\text{47.14g}$

$47.14 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * color(blue)("1 mL")/(color(purple)(3.102)color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("15.2 mL}}}}$

The answer is rounded to three sig figs, the number of significant figures you have for the number of moles of bromine present in the sample.