Answer:
[tex]\rho=1.54\ g/cm^3[/tex]
Explanation:
The expression for density is:
[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]
[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]
M is molar mass of Calcium = 40.078 g/mol
For cubic closest packed structure , Z= 4
[tex]\rho[/tex] is the density
Radius = 197 pm = [tex]1.97\times 10^{-8}\ cm[/tex]
Also, for fcc, [tex]Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm[/tex]
Thus,
[tex]\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3[/tex]
[tex]\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3[/tex]
[tex]\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3[/tex]
[tex]\rho=\frac{160.312}{104.19483}\ g/cm^3[/tex]
[tex]\rho=1.54\ g/cm^3[/tex]
