A) \[1.205\times {{10}^{-24}}\]atoms
B) \[1.205\times {{10}^{23}}\]atoms
C) \[1.205\times {{10}^{24}}\]atoms
D) \[1.205\times {{10}^{-23}}\]atoms
Correct Answer: C
Solution :
Number of atoms \[=\] Number of gram atoms \[\times \]Avogadro's number. Atomic weight of\[Cl=35.5\] Number of atoms of\[Cl=\] \[=\frac{71.00}{35.50}\times 6.023\times {{10}^{23}}\] \[=2\times 6.023\times {{10}^{23}}=1.025\times {{10}^{24}}\]
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