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

1 Answer
Oct 12, 2017

Answer:

#"5.7 g cm"^(-3)#

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.

http://99bricla01678.blogspot.ro/2013/05/calculating-density.html

So, you know that you mineral has a mass of #"262.2 g"# and a volume of #"46 cm"^3#. In order to find its density, #rho#, you need to divide its mass by its volume--this will give you the mass of exactly #"1 cm"^3# of this mineral.

#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.