What is the mass of a piece of metal that raises the volume of water in a graduated cylinder from #"15.0 cm"^3# to #"70.0 cm"^3# ? the density of iron is #"5.1 g/cm"^3#

1 Answer
Jul 26, 2016

Answer:

#"281 g"#

Explanation:

The idea here is that the volume of the metal will cause the volume of the water to increase from its initial value of #"15.0 cm"^3# to the final value of #"70.0 cm"^3#.

http://schools.birdville.k12.tx.us/cms/lib2/tx01000797/centricity/domain/912/chemlessons/Lessons/Density/Density.htm

In other words, the difference between the final volume of the water, i.e. the volume of the water + metal, and the initial volume of the water will give you the volume of the metal

#V_"metal" = "70.0 cm"^3 - "15.0 cm"^3 = "55.0 cm"^3#

Now, the problem tells you the metal has a density of #"5.1 g cm"^(-3)#. This means that every cubic centimeter of this metal has a mass of #"5.1 g"#.

In your case, #"55.0 cm"^3# will have a mass of

#55.0 color(red)(cancel(color(black)("cm"^3))) * overbrace("5.1 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(purple)("given density")) = color(green)(|bar(ul(color(white)(a/a)color(black)("281 g")color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.