Question

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

Answer 1

`"Density"`, `rho~=2*g*mL^-1`

Explanation:

By definition, `rho="Mass of substance"/"Volume of substance"`.

Here, we are given the mass and the volume with respective units of `"grams"` and `"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?

Answer 2

`"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 `"53 mL"`, which means that one unit of volume is `"1 mL"`. This implies that in order to find the density of the unknown substance, you must figure out the mass of `"1 mL"`.

You already know that `"53 mL"` have a mass of `"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 `"1 mL"` of this substance has a mass of `"2.1 g"`, you can say that its density is equal to

`color(darkgreen)(ul(color(black)("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.