# What is the approximate density of a mineral with a mass of 262.2 grams that displaces 46 cubic centimeters of water?

##### 1 Answer
May 21, 2017

$d = 5.7 {\text{ g"/"cm}}^{3}$

#### Explanation:

$d =$?
$m = 262.2 \text{ g}$

The volume of the mineral is equal to the volume of water it displaced, because the density of water is $1 {\text{ g"//"cm}}^{3}$
$V = 46 {\text{ cm}}^{3}$

Formula for density is

d=m/V=(262.2" g")/(46 " cm"^3)=5.7 " g"/"cm"^3