Gold has cubic crystals whose unic cell has edge length of 407.9 pm. Density of gold is 19.3 g `cm^(-3)`. Calculate the number of atoms per unit cell. Also predict the type of crystal lattice of gold (Atomic mass of gold = 197 amu)

Answer

4 fcc

Explanation

`"Density"(rho)=(zxxM)/(a^(3)xxN_(0)xx10^(-30))or z=(rhoxxa^(3)xxN_(0)xx10^(-30))/(M)`
Density of unit cell `(rho)=19.3"g cm"^(-3)` , Atomic mass of gold (M) = 197 amu = 197 `"g mol"^(-1)`
Avogadro's no. `(N_(0))=6.022xx10^(23)"mol"^(-1)` , Edge length of unit cell (a) = 407.9 pm = 407.9
`z=((19.3"g cm"^(-3))xx(407.9)^(3)xx(6.022xx10^(23)mol^(-1))xx(10^(-30)cm^(3)))/(("197 g" mol^(-1)))=4 ("fcc structure")`