A) \[10{{N}_{A}}\]
B) \[{{10}^{-1}}{{N}_{A}}\]
C) \[{{10}^{-2}}{{N}_{A}}\]
D) \[{{10}^{-3}}{{N}_{A}}\]
Correct Answer: D
Solution :
\[\because \]In 44 g of \[C{{O}_{2}}\], number of \[C{{O}_{2}}\] molecules = \[{{N}_{A}}\] \[\therefore \] In 44 mg of \[C{{O}_{2}}\], number of \[C{{O}_{2}}\] molecules \[={{N}_{A}}\times {{10}^{-3}}\] \[(1mg={{10}^{-3}}g)\]
You need to login to perform this action.
You will be redirected in
3 sec
