$\begin{array}{1 1}4.55\times 10^{-5}g\;CO\\3.55\times 10^{-5}g\;CO\\2.55\times 10^{-5}g\;CO\\1.55\times 10^{-5}g\;CO\end{array} $

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35 ppm CO,means $10^6g$ of air contains =$35g\;CO$

$\therefore$ Volume of air =$\large\frac{\text{mass of air}}{\text{density of air}}$

$\Rightarrow \large\frac{10^6}{1.3}$$=7.69\times 10^5L$

Now since $7.69\times 10^5L$ of air has =35g CO

$\therefore$ 1L of air has =$35/7.69\times 10^5\;CO$

$\Rightarrow 4.55\times 10^{-5}g\;CO$

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