Respuesta :
Answer:
Edge length = 0.40446 nm = [tex]4.0446\times 10^{-8}\ cm[/tex]
Volume = 0.06617 nm³ = [tex]6.617\times 10^{-23}\ cm^3[/tex]
[tex]\rho=2.708\ g/cm^3[/tex]
Explanation:
Given that:
The radius = 0.143 nm
For FCC,
[tex]Edge\ length=\frac {4}{\sqrt {2}}\times radius[/tex]
Thus,
[tex]Edge\ length=\frac{4}{\sqrt {2}}\times 0.143\ nm=2\sqrt{2}\times \:0.143\ nm=0.40446\ nm[/tex]
Edge length = 0.40446 nm
Also, 1 nm = [tex]10^{-7}[/tex] cm
So,
Edge length = [tex]4.0446\times 10^{-8}\ cm[/tex]
Volume = [tex]{(Edge\ length)}^3={0.40446}^3\ nm^3=0.06617\ nm^3[/tex]
Volume = 0.06617 nm³
Also, 1 nm³ = [tex]10^{-21}[/tex] cm³
Volume = [tex]6.617\times 10^{-23}\ cm^3[/tex]
The expression for density is:
[tex]\rho=\frac {Z\times M}{N_a\times {Volume}}[/tex]
[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]
M is molar mass = 26.98 g/mol
For FCC , Z= 4
Thus,
[tex]\rho=\frac {4\times 26.98\ g/mol}{6.023\times 10^{23}\ {mol}^{-1}\times 6.617\times 10^{-23}\ cm^3}[/tex]
[tex]\rho=\frac {107.92}{10^{-23}\times \:10^{23}\times \:6.023\times \:6.617}\ g/cm^3[/tex]
[tex]\rho=\frac{107.92}{39.854191}\ g/cm^3[/tex]
[tex]\rho=2.708\ g/cm^3[/tex]
