d(Cr) = 8,96 g/cm3
M(Cr) = 63,546 g/mol
In a cell face centred cubic, there are 4 atoms of Cur . . . !
n(Cr in 1 cm3) = m(Cu) / M(Cr)
n(Cr in 1 cm3) = 8,96 / 63,546
n(Cr in 1 cm3) = 0,141 mol of Cr
Avogadro's number is : NA = 6,02•10^23 mol^-1
N(Cr) = n(Cu) x NA
N(Cr) = 0,141 x 6,02•10^23
N(Cr) = 8,488•10^22 atoms of Cr
N(cell) = N(Cu) / 4
N(cell) = 8,488•10^22 / 4
N(cell) = 2,122•10^22 cells in 1 cm3
