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The ambient concentration of CO2 in a city is 500 ppm (v/v) on a particular day. Express its concentration as mg/L (ppm in m/v) of air. Assume STP. 0.223 g/L 0.500 g/L 0.761 g/L 0.981 g/L

Question

The ambient concentration of CO2 in a city is 500 ppm (v/v) on a particular day. Express its concentration as mg/L (ppm in m/v) of air. Assume STP. 0.223 g/L 0.500 g/L 0.761 g/L 0.981 g/L

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Solution

To convert the concentration of CO2 from ppm (v/v) to mg/L (ppm in m/v), we need to know the density of CO2 at standard temperature and pressure (STP). The density of CO2 at STP is approximately 1.98 g/L.

  1. First, convert ppm (v/v) to a fraction. 500 ppm = 500/1,000,000 = 0.0005 (v/v).
  2. Then, multiply this fraction by the density of CO2 to get the concentration in g/L. 0.0005 * 1.98 g/L = 0.00099 g/L.
  3. Finally, convert g/L to mg/L by multiplying by 1000 (since there are 1000 mg in a gram). 0.00099 g/L * 1000 = 0.99 mg/L.

So, the concentration of CO2 is approximately 0.99 mg/L. The closest answer to this is 0.981 g/L.

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