Explain how to solve Is this solution correct because here do not convert the$(44)$ to micro$-$gram because g over liter do not equal mg per meter cubic

a$)$ carbon dioxide $(C{O}_2),5000PPmV$

formul a

$m\frac{g}{m^3}=\frac{ppmv\text{m}\text{o}\text{l}\text{e}\text{w}\text{l}\text{a}\text{r}\text{w}\text{i}\text{g}\text{h}\text{t}\text{.}}{24\mathrm{.}465}$

$\frac{mg}{m^3}=\frac{5000(44)}{24\mathrm{.}465}$

$\frac{mg}{m^3}=8992$

b$)$ HCHO,$3\mathrm{.}6m\frac{g}{m^3}$

$m\frac{g}{m^3}=\frac{ppmV\text{m}\text{o}\text{l}\text{c}\text{u}\text{l}\text{a}\text{r}\text{n}\text{e}\text{i}\text{g}\text{h}\text{t}\text{.}}{24\mathrm{.}46(s)}$

$ppmV=\frac{24\mathrm{.}565m\frac{g}{m^3}}{\text{m}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{.}}$

$PpMV=\frac{24\mathrm{.}563\mathrm{.}6}{(2+12+16)}$

$PpMv=2\mathrm{.}94$

c$)$

$No,25PpMV$

$m\frac{g}{m^3}=\frac{Ppm\text{m}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{v}\text{e}\text{i}\text{g}\text{h}\text{t}\text{.}}{24\mathrm{.}465}$

$=25(14+16)$

$m\frac{g}{m^3}={30\mathrm{.}77}^{24\mathrm{.}465}$

If any doupt then please commert. If any Information regarding $PPm$ and $PPMV$ them ask.