Question

The specific gravity of iron is `7.87` and the density of water at `4.00^@"C"` is `"1.00 g/cm"^3`. You can use this information to find the density of iron. Find the volume occupied by `"9.50 g"` of iron?

Answer 1

Here's what I got.

Explanation:

The idea here is that the specific gravity of a substance is calculated by dividing the density of that substance by the density of a reference substance, which is usually water at `4^@"C"`.

`"SG" = rho_"substance"/rho_ ("H"_2"O at 4"^@"C")`

This means that in order to find the density of a substance, you must multiply its specific gravity by the density of the reference substance.

`rho_"Fe" = "SG" * rho_ ("H"_ 2"O at 4"^@"C")`

Since you are told that water at `4^@"C"` has a density of `"1.00 g cm"^(-3)` and that iron has a specific gravity of `7.87`, you can say that its density must be equal to

`rho_ "Fe" = 7.87 * "1/00 g cm"^(-3)`

`rho_ "Fe" = "7.87 g cm"^(-3)`

Now, the density of a substance tells you the mass of exactly `1` unit of volume of that substance. In this case, the density of iron tells you that `:1 cm"^3` of iron has a mass of `"7.87 g"`.

This means that a `"9.50-g"` sample of iron will occupy a volume of

`9.50 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(7.87color(red)(cancel(color(black)("g")))) = "1.21 cm"^3`

The answer is rounded to three sig figs.