Density of Li atom is 0.53 g//cm^3 . The edge length of Li is 3.5 Å.Find out the number of Li atoms in a unit cell (N_0=6.023xx10^23 , M=6.94)

Density of Li atom is 0.53 g//cm^3 . The edge length of Li is 3.5 Å.F

1.	Density of Li atom is 0.53 g//cm^3 . The edge length of Li is 3.5 Å.Find out the number of Li atoms in a unit cell (N_0=6.023xx10^23 , M=6.94)
Answer» SOLUTION :The aim is to FIND Z in the FORMULA `rho=(ZxxM)/(a^3xxN_0)` `therefore Z=(rhoxxa^3xxN_0)/M=(0.53 "G cm"^(-3)XX(3.5xx10^(-8) cm)^3xx(6.023xx10^3 "mol"^(-1)))/(6.94 "g mol"^(-1))=1.97 =2`