# 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

Jul 26, 2016

$\text{281 g}$

#### 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 ${\text{15.0 cm}}^{3}$ to the final value of ${\text{70.0 cm}}^{3}$. 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

${V}_{\text{metal" = "70.0 cm"^3 - "15.0 cm"^3 = "55.0 cm}}^{3}$

Now, the problem tells you the metal has a density of ${\text{5.1 g cm}}^{- 3}$. This means that every cubic centimeter of this metal has a mass of $\text{5.1 g}$.

In your case, ${\text{55.0 cm}}^{3}$ 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.