A) 7.3
B) 8.1
C) 5.1
D) 11.3
Correct Answer: A
Solution :
In bcc unit cell, \[4r=\sqrt{3a}\]\[\Rightarrow \,\,r(Cr)=\frac{\sqrt{3}a}{4}=\frac{\sqrt{3}}{4}\times 287pm\] |
\[=124.3pm\] |
N = number of atoms per unit cell, |
M = molar mass |
\[{{a}^{3}}\] = volume of cubic unit cell, |
\[{{N}_{A}}\] = Avogadro's number |
For bcc Z = 2 |
\[=\frac{2\times 52g}{6.023\times {{10}^{23}}}\times {{\left( \frac{1}{2.87\times {{10}^{-8}}cm} \right)}^{3}}\] \[=7.3g/c{{m}^{3}}\] |
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