A) \[6.8\times {{10}^{23}}\]atoms
B) \[8.8\times {{10}^{17}}\]atoms
C) \[8.8\times {{10}^{22}}\]atoms
D) \[6.9\times {{10}^{17}}\]atoms
Correct Answer: D
Solution :
Atomic weight of\[Se=78.96\,\,g\] Number of moles in \[90\] micrograms \[=\frac{90\times {{10}^{-6}}}{78.96}=1.139\times {{10}^{-6}}\]moles \[1\] mole of selemium atoms\[=6.023\times {{10}^{23}}\] \[1.139\times {{10}^{-6}}\]moles of selemium atoms\[=1.139\times {{10}^{-6}}\times 6.023\times {{10}^{23}}\] \[=6.9\times {{10}^{17}}\]atoms
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