A) \[1.66\times {{10}^{-23}}c{{m}^{3}}\]
B) \[1.66\times {{10}^{-20}}c{{m}^{3}}\]
C) \[1.66\times {{10}^{-22}}c{{m}^{3}}\]
D) \[1.66\times {{10}^{-18}}c{{m}^{3}}\]
Correct Answer: A
Solution :
Because the number of atom in \[108g\] \[Ag=6.023\times {{10}^{23}}\] The number of atom in \[10.8g\] \[Ag=6.023\times {{10}^{22}}\] because, the value of \[10.8g\text{ }Ag=1c{{m}^{3}}\] \[\therefore \]the volume of \[6.023\times {{10}^{22}}\]atom\[=1c{{m}^{3}}\] \[\therefore \] the volume of 1 atom \[=\frac{1}{6.023\times {{10}^{22}}}c{{m}^{3}}\] \[=1.66\times {{10}^{-23}}c{{m}^{3}}\]
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