$\begin{array}{1 1}(a)\;0.10mL\;and\;3.01\times 10^{22}\\ (b)\;0.20mL\;and\;3.01\times 10^{21} \\(c)\;0.20mL\;and\;6.02\times 10^{23}\\(d)\;0.10mL\;and\;6.02\times 10^{21}\end{array} $

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At STP 22400ml of $NO_2$=46g of $NO_2$

$112.0$ml of $NO_2=\large\frac{112.0mL\times 46.0g}{22400mL}$$=0.23$g

$V_{NO_2}=\large\frac{Mass}{Density}=\frac{0.23}{1.15gmL^{-1}}$$=0.20mL$

Number of molecules =$\large\frac{0.23}{46}$$\times 6.02\times 10^{23}$

$\Rightarrow 3.01\times 10^{21}$

Hence (b) is the correct answer.

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