A) \[2.57\times {{10}^{21}}\] unit cells
B) \[5.14\times {{10}^{21}}\] unit cells
C) \[1.28\times {{10}^{21}}\] unit cells
D) \[1.71\times {{10}^{21}}\] unit cells
Correct Answer: A
Solution :
\[\frac{1}{58.5}\times 6.023\times {{10}^{23}}=1.029\times {{10}^{22}}\] A unit cell contains \[4\,N{{a}^{+}}\]ion and \[4\,C{{l}^{-}}\]ions \Unit cell \[=\frac{1.029\times {{10}^{22}}}{4}\]\[=2.57\times {{10}^{21}}\]unit cell.
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