Answer:
[tex]\rho=8907.94\ Kg/m^3[/tex]
Explanation:
Given that
a=3.524 A
At.Wt. ,M= 58.7 g/mole,
For FCC
Z = 4
[tex]4r=\sqrt2\ a[/tex]
The density given as
[tex]\rho=\dfrac{ZM}{N_Aa^3}[/tex]
[tex]\rho=\dfrac{4\times 58.7\times 10^{-3} }{ 6.023\times 10^{23}\times (3.524\times 10^{-10})^3}[/tex]
[tex]\rho=8907.94\ Kg/m^3[/tex]
So the density is [tex]\rho=8907.94\ Kg/m^3[/tex]