# What is the approximate density of the mineral with a mass of #"262.2 g"# that displaces #"46 cm"^3# of water?

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that the volume of water **displaced** by the mineral will actually give you the **volume** of the mineral.

When you place the mineral in a graduated cylinder, for example, that contains a known volume of water, the level of the water will **rise**.

The *difference* between the level of the water **with** the mineral and the level of the water *without* the mineral is the volume of water **displaced** by the mineral, i.e. the volume of the mineral.

So, you know that you mineral has a mass of **density**,

#rho = "262.2 g"/"46 cm"^3 = color(darkgreen)(ul(color(black)("5.7 g cm"^(-3))))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the volume of the water displaced by the mineral.