# What is the density of a substance that has a mass of 112 g and a volume of 53 mL?

Jun 10, 2017

$\text{Density}$, $\rho \cong 2 \cdot g \cdot m {L}^{-} 1$

#### Explanation:

By definition, $\rho = \text{Mass of substance"/"Volume of substance}$.

Here, we are given the mass and the volume with respective units of $\text{grams}$ and $\text{millilitres}$, and we can thus directly assess the quotient without pfaffing about with units.........

rho_"substance"=(112*g)/(53*mL)=??*g*mL^-1.

Would this substance float? Why or why not?

Jun 10, 2017

${\text{2.1 g mL}}^{- 1}$

#### Explanation:

Density is simply a measure of the mass of exactly one unit of volume of a given substance.

In this case, your sample is said to occupy a volume of $\text{53 mL}$, which means that one unit of volume is $\text{1 mL}$. This implies that in order to find the density of the unknown substance, you must figure out the mass of $\text{1 mL}$.

You already know that $\text{53 mL}$ have a mass of $\text{112 g}$, so set up the equation like this

overbrace((color(blue)(?)color(white)(.)"g")/"1 mL")^(color(red)("the density of the substance")) = overbrace("112 g"/"53 mL")^(color(red)("its known composition"))

Rearrange and solve to find

color(blue)(?) = (1 color(red)(cancel(color(black)("mL"))))/(53color(red)(cancel(color(black)("mL")))) * "112 g"

color(blue)(?) ="2.1 g"

So, if $\text{1 mL}$ of this substance has a mass of $\text{2.1 g}$, you can say that its density is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{density = 2.1 g mL}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the sample.