If carbon dioxide expresses a $1"ppm"$ concentration in water, what is its concentration in $molL^-1$ ?

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Answer 1 · 2017-03-25T07:20:29

Well $"1 ppm "-=" 1 mg"*L^-1............$

And so you simply multiply the $"ppm"$ concentration by $(1xx10^-3*g*L^-1)/(44.0*g*mol^-1)$ to give a molar concentration in $mol*L^-1$ .

If carbon dioxide expresses a 1*"ppm"1*"ppm" concentration in water, what is its concentration in mol*L^-1mol*L^-1?