Answer:
[tex]\rho=7.15\ g/cm^3[/tex]
Explanation:
The expression for density is:
[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]
[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]
M is molar mass of Chromium = 51.9961 g/mol
For body-centered cubic unit cell , Z= 2
[tex]\rho[/tex] is the density
Radius = 125 pm = [tex]1.25\times 10^{-8}\ cm[/tex]
Also, for BCC, [tex]Edge\ length=\frac{4}{\sqrt{3}}\times radius=\frac{4}{\sqrt{3}}\times 1.25\times 10^{-8}\ cm=2.89\times 10^{-8}\ cm[/tex]
Thus,
[tex]\rho=\frac{2\times \:51.9961}{6.023\times \:10^{23}\times \left(2.89\times 10^{-8}\right)^3}\ g/cm^3[/tex]
[tex]\rho=7.15\ g/cm^3[/tex]
The density of solid crystalline chromium will be "7.15 g/cm³".
According to the question,
Radius,
Molar mass of Chromium,
For body-centered,
For BCC,
The edge length will be:
= [tex]\frac{4}{\sqrt{3} }\times radius[/tex]
= [tex]\frac{4}{\sqrt{3} }\times 1.25\times 10^{-8}[/tex]
= [tex]2.89\times 10^{-8} \ cm[/tex]
hence
The expression for density will be:
→ [tex]\rho = \frac{Z\times M}{N_a(Edge \ length)^3}[/tex]
By substituting the values, we get
[tex]= \frac{2\times 51.9961}{6.023\times 10^{23}\times (2.89\times 10^{-8})}[/tex]
[tex]= 7.15 \ g/cm^3[/tex]
Thus the above approach is right.
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