A) \[6.023\times {{10}^{23}}\]
B) \[6.023\times {{10}^{22}}\]
C) \[12.046\times {{10}^{23}}\]
D) none of these
Correct Answer: C
Solution :
Key Idea: Mol. wt. in gram \[=6.023\times {{10}^{23}}\] molecules Molecular weight of \[C{{O}_{2}}=12+2+16=44\] \[\therefore \] 44 g of CO 2 will have \[=6.023\times {{10}^{23}}\]molecules of \[C{{O}_{2}}\] \[\because \] One \[C{{O}_{2}}\] molecule has two oxygen atoms \[\therefore \] Number of oxygen atoms in 44 g of \[C{{O}_{2}}\] \[=2\times 6.023\times {{10}^{23}}\] \[=12.046\times {{10}^{23}}\]
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