# Question 228b4

Sep 8, 2017

${\text{100 cm}}^{3}$

#### Explanation:

The thing to remember about the density of a substance is that it's supposed to tell you the mass of exactly $1$ unit of volume of that substance.

In your case, the boulder is said to have a density of "8 g/"color(blue)("cm"^(3))#, which means that $\textcolor{b l u e}{{\text{1 cm}}^{3}}$ of this boulder has a mass of $\text{8 g}$. In other words, every time you have a mass of $\text{8 g}$ of this boulder, the sample will occupy $\textcolor{b l u e}{{\text{1 cm}}^{3}}$.

You can thus use the density of the boulder as a conversion factor to calculate the volume of the $\text{800-g}$ sample

$800 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g"))) * overbrace(color(blue)("1 cm"^3)/(8color(red)(cancel(color(black)("g")))))^("the given density = 8 g/"color(blue)("cm"^3)) = color(darkgreen)(ul(color(black)("100 cm}}^{3}}}}$

The answer is rounded to one significant figure.

So remember, when a problem provides you with the density of the substance, it's essentially giving you the mass of exactly $1$ unit of volume of that substance.