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

Well $\text{1 ppm "-=" 1 mg} \cdot {L}^{-} 1. \ldots \ldots \ldots . .$
And so you simply multiply the $\text{ppm}$ concentration by $\frac{1 \times {10}^{-} 3 \cdot g \cdot {L}^{-} 1}{44.0 \cdot g \cdot m o {l}^{-} 1}$ to give a molar concentration in $m o l \cdot {L}^{-} 1$.