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Chapter 18: Problem 15

The average concentration of carbon monoxide in air in an Ohio city in 2006 was 3.5 ppm. Calculate the number of CO molecules in \(1.0 \mathrm{~L}\) of this air at a pressure of 759 torr and a temperature of \(22^{\circ} \mathrm{C}\).

Short Answer

Expert verified
There are approximately \(8.61 \times 10^{16}\) CO molecules in 1.0 L of the air sample in this Ohio city.

Step by step solution

01

Convert temperature to Kelvin

First, we need to convert the given temperature of 22°C to Kelvin since all calculations in gas laws are done in Kelvin. To convert from Celsius to Kelvin, we add 273.15. Temperature in Kelvin = Temperature in Celsius + 273.15 Temperature in Kelvin = 22 + 273.15 Temperature in Kelvin = 295.15 K
02

Convert pressure to atm

Next, we need to convert the given pressure of 759 torr to atmospheres (atm) since the Ideal Gas Law uses pressures in atm. We can do this using the conversion factor 1 atm = 760 torr. Pressure in atm = Pressure in torr / (760 torr/atm) = 759 torr / 760 torr/atm = 0.9987 atm
03

Calculate the moles of air

We will use the Ideal Gas Law (PV=nRT) to calculate the moles of air (n_air) in the given volume, pressure, and temperature. The gas constant (R) is 0.08206 L*atm/mol*K. n_air = PV / RT n_air = (0.9987 atm)(1.0 L) / (0.08206 L*atm/mol*K)(295.15 K) n_air = 0.04090 mol
04

Calculate the moles of CO

Now that we have the moles of air, we can use the ppm concentration to determine the moles of CO (n_CO). n_CO = 3.5 ppm * n_air = (3.5 * 10^(-6)) * 0.04090 mol n_CO = 1.4315 * 10^(-7) mol
05

Calculate the number of CO molecules

Finally, we can calculate the number of CO molecules by multiplying the moles of CO by Avogadro's number (6.022 × 10^23 molecules/mol). CO molecules = n_CO * Avogadro's number CO molecules = (1.4315 * 10^(-7) mol)(6.022 × 10^23 molecules/mol) CO molecules ≈ 8.61 × 10^16 There are approximately \(8.61 \times 10^{16}\) CO molecules in 1.0 L of the air sample in this Ohio city.

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