A) \[6\times {{10}^{20}}\]
B) \[3\times {{10}^{22}}\]
C) \[1.5\times {{10}^{23}}\]
D) \[0.5\times {{10}^{24}}\]
Correct Answer: C
Solution :
\[58.5\,g\,\,NaCl=1\,mole\]\[=6.02\times {{10}^{23}}N{{a}^{+}}C{{l}^{-}}units\]. One unit cell contains \[4\,N{{a}^{+}}C{{l}^{-}}\] 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.
You will be redirected in
3 sec
