silver from a a CCP letters the the edge length of its unit cell is 408.6 picometre calculate the density of a silver NA= 6.022× 10restiu 23 mol_1 atomic mass 108 g/ mol
Answers
Answered by
1
Answer:
The edge length of it's unit cel is 408.6 picometre
or
408.6 × 10^-10cm
The volume of the unit cel =a^3
=(408.6 × 10^-10cm)^3
= 68.27 × 10^-24 cm^3
ccp unit cells has 4 atoms per unit cell
Mass of one Ag atom
= molar mass of Ag /a vogadros number
= 108g/mol /6.022× 10^23 /mol
= 17.96×10^23g
Mass of unit cell =4×17.93 ×10^23g
=71.72×10^23g
Density of silver = mass of unit cell/volume of unit cell
Density of silver = 71.72×10^23g/68.27×10^-24cm^3
= 10.51g/cm^3.
I hope it's help a lot.
Similar questions