A) \[27\,\,g\,\,mo{{l}^{-1}}\]
B) \[20\text{ }g\text{ }mo{{l}^{-1}}\]
C) \[40\text{ }g\text{ }mo{{l}^{-1}}\]
D) \[30\text{ }g\text{ }mo{{l}^{-1}}\]
Correct Answer: A
Solution :
\[d=\frac{ZM}{{{N}_{A}}{{a}^{3}}}\] (Z=4 for fcc) \[M=\frac{d\times {{N}_{A}}\times {{a}^{3}}}{Z}\] \[=\frac{2.72\times 6.023\times {{10}^{23}}\times {{(404\times {{10}^{-10}})}^{3}}}{4}\] \[M=26.99\approx 27g\,mo{{l}^{-1}}\]You need to login to perform this action.
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