**Step 1: Calculate volume of the unit cell from the density**

$\mathbf{volume}\mathbf{=}\frac{\mathbf{47}\mathbf{.}\mathbf{867}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Ti}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Ti}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Ti}}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times}{\mathbf{10}}^{\mathbf{23}}\mathbf{}\overline{)\mathbf{Ti}\mathbf{}\mathbf{atoms}}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\overline{)\mathbf{Ti}\mathbf{}\mathbf{atoms}}}{\mathbf{1}\mathbf{}\mathbf{unit}\mathbf{}\mathbf{cell}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}{\mathbf{cm}}^{\mathbf{3}}}{\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Ti}}}$

**volume = 3.5328 x10 ^{-23} cm^{3}/1 unit cell**

**Step 2: Calculate the edge length of the unit cell using the volume**

$\mathbf{volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{unit}\mathbf{}\mathbf{cell}\mathbf{=}\mathbf{volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{a}\mathbf{}\mathbf{cube}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{volume}}{\mathbf{}}{\mathbf{of}}{\mathbf{}}{\mathbf{unit}}{\mathbf{}}{\mathbf{cell}}{\mathbf{=}}{{\mathbf{a}}}^{{\mathbf{3}}}}$

Titanium metal has a body-centered cubic unit cell. The density of titanium is 4.50 g/cm^{3}. Calculate the edge length of the unit cell and a value for the atomic radius of titanium. ( *Hint*: In a body-centered arrangement of spheres, the spheres touch across the body diagonal.)

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.