A) \[3.01\times {{10}^{20}}\]
B) \[3.01\times {{10}^{12}}\]
C) \[3.01\times {{10}^{23}}\]
D) \[3.01\times {{10}^{24}}\]
Correct Answer: B
Solution :
\[\because \] 22400cc of gas at STP has \[6\times {{10}^{23}}\]molecules \[\therefore \] 1.12 \[\times {{10}^{-7}}\] of gas at STP has \[\frac{6\times {{10}^{23}}\times 1.12\times {{10}^{-7}}}{22400}\] \[=.03\times {{10}^{14}}=3\times {{10}^{12}}\].
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