A) \[6\times {{10}^{20}}\]
B) \[1.5\times {{10}^{23}}\]
C) \[3\times {{10}^{22}}\]
D) \[0.5\times {{10}^{24}}\]
Correct Answer: B
Solution :
[b] 58.5 g \[NaCl=1mol=6.02\times {{10}^{23}}N{{a}^{+}}C{{l}^{-}}\]units One unit cell contains \[4N{{a}^{+}}C{{l}^{-}}\]units. Hence number of unit cell present \[=\frac{6.02\times {{10}^{23}}}{4}=1.5\times {{10}^{23}}\]You need to login to perform this action.
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