We’re asked to **determine how many atoms are present in a cube of silver** metal that **measures 1.050 cm on each edge **given its density.

**We can calculate the # of Ag (silver) atoms using the following steps:**

**Step 1: **Determine the** volume of the silver cube**. Recall that the * volume of a cube* is given by the equation:

$\overline{){\mathbf{Volume}}{\mathbf{=}}{{\mathbf{l}}}^{{\mathbf{3}}}}$

*l = length of edge of a cube*

**Step 2: **Determine the** mass of the silver cube** from the given density from the equation:

$\overline{){\mathbf{density}}{\mathbf{=}}\frac{\mathbf{mass}}{\mathbf{volume}}}\phantom{\rule{0ex}{0ex}}$

This means that: **Mass = density x volume**

**Step 3**: Determine atoms Ag. The flow of the solution will be like this:

** mass Ag (molar mass Ag) ** → moles Ag

You are given a cube of silver metal that measures 1.050 cm on each edge. The density of silver is 10.5 g/cm^{3}.

How many atoms are in this cube?