# The density of gold is 19.3 g/mL. What volume (in mL) will a 1.592 lb sample of gold occupy?

Volume occupied is approx. $38 \cdot m L$
$\text{Density } \rho$ $=$ $\text{Mass"/"Volume}$
And thus $\text{Volume}$ $=$ $\frac{\text{Mass}}{\rho}$ $=$ $\frac{1.592 \cdot l b \times 454 \cdot g \cdot l {b}^{-} 1}{19.3 \cdot g \cdot m {L}^{-} 1}$ $=$ ??*mL
It doesn't seem a very big nugget does it? Nevertheless, the nugget would be worth over \$30000*USD.