A) \[3.01\times {{10}^{12}}\]
B) \[3.01\times {{10}^{24}}\]
C) \[3.01\times {{10}^{23}}\]
D) \[3.01\times {{10}^{20}}\]
Correct Answer: A
Solution :
Given, \[V=1.12\times {{10}^{-7}}c{{m}^{3}}\] \[\because \]\[22400\text{ }c{{m}^{3}}\]of the gas at STP has molecules \[=6.02\times {{10}^{23}}\] \[\therefore \]\[1.12\times {{10}^{-7}}c{{m}^{3}}\]of the gas at STP will have molecules \[=\frac{6.02\times {{10}^{23}}}{22400}\times 1.12\times {{10}^{-7}}\] \[=3.01\times {{10}^{12}}\]molecules
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