# 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?

Jan 22, 2018

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 ${4}^{\circ} \text{C}$.

"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 ${4}^{\circ} \text{C}$ has a density of ${\text{1.00 g cm}}^{- 3}$ and that iron has a specific gravity of $7.87$, you can say that its density must be equal to

${\rho}_{\text{Fe" = 7.87 * "1/00 g cm}}^{- 3}$

${\rho}_{\text{Fe" = "7.87 g cm}}^{- 3}$

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 $\text{7.87 g}$.

This means that a $\text{9.50-g}$ 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.