A) \[7.94\,\,g\,\,m{{L}^{-1}}\]
B) \[8.96\,\,g\,\,m{{L}^{-1}}\]
C) \[2.78\,\,g\,\,m{{L}^{-1}}\]
D) \[6.72\,\,g\,\,m{{L}^{-1}}\]
Correct Answer: A
Solution :
\[d=\frac{ZM}{{{N}_{A}}{{a}^{3}}}(for\,\,bcc,\,\,Z=2)\] \[{{d}_{Fe}}=\frac{(2)\times 56.0\,\,g\,\,mo{{l}^{-1}}}{(6.02\times {{10}^{23}}mo{{l}^{-1}}){{(2.861\times {{10}^{-8}})}^{3}}c{{m}^{3}}}\] \[=7.94\,\,g\,\,m{{L}^{-1}}\]
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