Question

Silver has a density of $10.5\;g\,c{m^{ - 3}}$ and gold has a density of $19.3\;g\,c{m^{ - 3}}$. Which would have a greater mass, $5\;E\,c{m^{ - 3}}$ of silver or $5\;c{m^3}$ of gold?

Answer 1

The density of any substance can be defined as the mass of the substance per unit of its volume. Most often density is denoted as $\rho$, the lower-case Greek letter $rho$, we can also use the Latin letter $D$. Mathematically, density is the mass divided by volume.
Formula used:
$\rho = \dfrac{m}{V}$
Where $\rho$is the density, $m$ is the mass of the substance and $V$ is the volume of the substance.

Complete answer:
To answer this question, we have to first find the density of both elements i.e., silver and gold. According to the definition density is the mass per unit volume so the density of both elements are as follows.
$Density = \dfrac{{Mass}}{{Volume}}$
Thus, Mass of silver is,
$\Rightarrow mass = density\left( \rho \right) \times volume$
$\Rightarrow mass\,of\;silver = 10.5\;g\,c{m^{ - 3}} \times 5\;c{m^3}$
$\Rightarrow 52.5\;g$
Similarly, mass of gold is,
$\Rightarrow mass = density\left( \rho \right) \times volume$
$mass\,of\,gold = \;19.3\,g\;c{m^{ - 3}} \times 5\,c{m^3}$
$\approx 100\,g$
We can see that the mass of gold is more than the mass of silver i.e., $100\;g$.

Addition information:
Gold is one of the precious metals because of its physical and chemical properties. It has the greatest ductility of all metals, and also the most malleable. It is also the most versatile of all the metals. One of the most valuable properties is its density. Density is termed as the amount of mass of substance per unit of its volume. Simply,
$density = \dfrac{{mass}}{{volume}}$

Note:
Here in this problem, we answer in grams which come out after cancelling the units. Make sure to check whether there is any mistake in division, multiplication, addition correctly in your solution, it is also making errors very easily.