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 :
| [a] \[\frac{1}{58.5}\times 6.023\times {{10}^{23}}=1.029\times {{10}^{22}}\] |
| A unit cell contains \[4N{{a}^{+}}\] ion and \[4C{{l}^{-}}\] ions |
| \[\therefore \] Unit cell\[\text{=}\frac{1.029\times {{10}^{22}}}{4}=2.57\times {{10}^{21}}\] unit cell |
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