Question

Initially 10.0 mL of water was placed in a graduate cylinder and then a solid metal was immersed in the water in the graduated cylinder. At this point, the volume of water was read as 15.5 mL. If the mass of the solid is 9.25 g, what is the density?

Answer 1

`"1.85 g mL"^(-1)`

Explanation:

The idea here is that the increase in the volume of water in the cylinder is due to the volume of the metal.

This means that you can figure out the volume of the metal by looking at the volume occupied by the water + metal and at the volume occupied by the water alone in the cylinder.

More specifically, the volume of the metal will be

`V_"metal" = V_"water + metal" - V_"water"`

In your case, you will have

`V_"metal" = "15.5 mL" - "10.0 mL" = "5.0 mL"`

Now, the density of the metal tells you the mass of one unit of volume of this metal. Since you're working with milliliters, one unit of volume will be `"1 mL"`.

Your goal will thus be to figure out the mass of `"1 mL"` of this metal. You know that `"5.0 mL"` have a mass of `"9.25 g"`, so use this as a conversion factor to find

`1.0 color(red)(cancel(color(black)("mL"))) * "9.25 g"/(5.0color(red)(cancel(color(black)("mL")))) = "1.85 g"`

So, if `"1 mL"` has amass of `"1.85 g"`, it follows that the density of the metal will be

`"density" = color(green)(|bar(ul(color(white)(a/a)color(black)("1.85 g mL"^(-1))color(white)(a/a)|)))`

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