we have give that aluminium forms Fcc cubic crystal.
density = 2.7g cm-3
Mass of aluminium = 27g
number of unit cell =4 (FCC)
we have find edge length
Thus using formula
length of the unit cell edge = 400Pm =400x10-10 cm
volume of the unit cell = ( 400 x10-10cm)3
= 6.4 x 10-23 cm-3
as the element A forms a body centred cubic lattice so no of atom per units cell is 2
z= 2 atoms unit cell
atomic mass of the element =100 g/mol
density of element is given by
we have given,.
mass =108 g
edge (a) = 409pm
a3 = 6.84 x1023 cm3
Na =6.023 x10-233
Z =4
Apply formula
density = 10.98 g cm–3
we know that
as given that,
density= 2.75 g cm–3
a = 654 pm or a3 = 2.79 x 10-23
mass of KBr is 119 gram
we have given :
mass of silver = 107.87 g
Z(FCC) = 4
distance between nearest neighbour Ag atoms = 2.87 x10-10