At 1 atm and 300 K, air contains 340 ppmv $CO_2$. Assuming ideal gases, find its concentration in mg/m³,to two decimals. Use molecular weight 44 g/mol and $R=8.314\,\mathrm{J/(mol\,K)}$.
The mole fraction equals the volume fraction,$x=340\times10^{-6}$. Total molar density is $P/(RT)$,with $P=101325$ Pa. Therefore $$C=x\frac{P}{RT}M(1000)=340\times10^{-6}\frac{101325}{8.314(300)}44(1000)=607.73875\ldots\ \mathrm{mg/m^3}.$$ The factor 1000 converts grams to milligrams. Answer **607.74 mg/m³**.
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