A) \[6.023\times {{10}^{21}}\]molecules of \[C{{O}_{2}}\]
B) 22.4 L of \[C{{O}_{2}}\]at STP
C) 0.44 g of \[C{{O}_{2}}\]
D) None of these
Correct Answer: B
Solution :
a. \[6.023\times {{10}^{23}}\]molecules of \[C{{O}_{2}}\] No. of atoms \[=3\times 6.023\times {{10}^{21}}\]=\[18.069\times {{10}^{21}}\] atoms b. 22.4L of \[C{{O}_{2}}\] No. of atoms =\[6.023\times {{10}^{23}}\times 3\]\[=18.069\times {{10}^{23}}\]atoms c. 0.44gm of \[C{{O}_{2}}\] No. of moles \[=\frac{0.44}{44}=\frac{1}{100}\times 6.023\times {{10}^{23}}\]moles \[=6.023\times {{10}^{21}}\]moles \[=3\times 6.023\times {{10}^{21}}\]atoms \[18.069\times {{10}^{21}}\]atoms
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