A) \[3.2\times {{10}^{6}}\]
B) \[3.2\times {{10}^{10}}\]
C) \[3.2\times {{10}^{13}}\]
D) \[3.2\times {{10}^{4}}\]
Correct Answer: C
Solution :
According to question, \[{{P}_{1}}={{10}^{-6}}mm\,Hg\] \[{{P}_{2}}=760\,mm\,Hg\] (S.T.P. conditions) \[{{T}_{1}}=298\,K\] \[{{T}_{2}}=273\,K\] \[{{V}_{1}}=1\,\,litre\] \[{{V}_{2}}=?\] From gas equation, \[\frac{{{P}_{1}}\,{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}\,{{V}_{2}}}{{{T}_{2}}}\] or \[{{V}_{2}}=\frac{{{10}^{-6}}\times 1\times 273}{760\times 298}=1.205\times {{10}^{-9}}\,\text{litre}\] Hence, number of molecules in \[1.205\times {{10}^{-9}}\] litre gas at S.T.P. \[=\frac{6.023\times {{10}^{23}}\times 1.205\times {{10}^{-9}}}{22.4}\] \[=3.24\times {{10}^{13}}\]
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