# Question e0d85

Sep 1, 2015

The density of the gold bar is ${\text{17.1 g/cm}}^{3}$.

#### Explanation:

This time, you know the mass of the gold bar, and you can use its dimensions to determine its volume, which means that you can calculate its density.

Again, treat the gold bar as a rectangular prism of dimensions $l$ by $w$ by $h$

$V = l \times w \times h$

$V = {\text{17.18 cm" xx "9.21 cm" xx "4.45 cm" = "728.7 cm}}^{3}$

Since the mass was given to you in kilograms, but you need to express the density in grams per cubic centimeter, convert the mass to grams first

12.441color(red)(cancel(color(black)("kg"))) * "1000 g"/(1color(red)(cancel(color(black)("kg")))) = "12441 g"

The density will thus be

$\rho = \frac{m}{V}$

rho = "12441 g"/("728.7 cm"""^3) = color(green)("17.1 g/cm"""^3)#

Since the density of the gold bar is smaller than the density of pure gold, you can say for sure that you're dealing with impure gold. That happens because the metals that are usually mixed with gold, silver, nickel, copper, stuff like that, have smaller densities than pure gold.

As a result, the mixture will have a smaller density when compared with the same mass of pure gold, a heavier metal.