# Problem: You are given a cube of silver metal that measures 1.050 cm on each edge. The density of silver is 10.5 g/cm3.Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

###### FREE Expert Solution

$\overline{){\mathbf{V}}{\mathbf{=}}\frac{\mathbf{4}}{\mathbf{3}}{{\mathbf{\pi r}}}^{{\mathbf{3}}}}$

$\frac{\mathbf{V}}{\frac{\mathbf{4}}{\mathbf{3}}\mathbf{\pi }}{\mathbf{=}}\frac{\overline{)\frac{\mathbf{4}}{\mathbf{3}}\mathbf{\pi }}{\mathbf{r}}^{{\mathbf{3}}}}{\overline{)\frac{\mathbf{4}}{\mathbf{3}}\mathbf{\pi }}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt[\mathbf{3}]{\frac{\mathbf{3}}{\mathbf{4}}\mathbf{\pi V}}{\mathbf{=}}\sqrt[\mathbf{3}]{{\mathbf{r}}^{\mathbf{3}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{r}}{\mathbf{=}}\sqrt[\mathbf{3}]{\frac{\mathbf{3}}{\mathbf{4}}\mathbf{\pi V}}}$

From the previous calculation, the volume of a single silver atom is 1.09×10-23 cm3.

###### Problem Details

You are given a cube of silver metal that measures 1.050 cm on each edge. The density of silver is 10.5 g/cm3.

Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.