The average concentration of carbon monoxide in air in a city in 2007 was 3.0 ppm. Calculate the number of CO molecules in \(1.0 \mathrm{~L}\) of this air at a pressure of \(100 \mathrm{kPa}\) and a temperature of \(25^{\circ} \mathrm{C}\).

The concentration of CO in the air is given as 3.0 ppm (parts per million). We need to convert this to a fraction to use it in our calculations. Since there are one million parts in "parts per million", we have a ratio of CO molecules to total molecules: \[ \frac{3.0}{1,000,000} \]

The ideal gas law equation is: \[P V = n R T\] where \(P\) is the pressure, \(V\) is the volume, \(n\) is the amount in moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. We are given the pressure \(P = 100 \mathrm{kPa}\), the volume \(V = 1.0 \mathrm{~L}\), and the temperature \(T = 25^{\circ} \mathrm{C}\). First, we need to convert temperature to Kelvin: \[T = 25^{\circ} \mathrm{C} + 273.15 = 298.15\mathrm{K}\] And the given pressure should be converted to atm: \[P = \frac{100 \mathrm{kPa}}{101.325} = 0.9869 \mathrm{atm}\] Now, the value of the gas constant \(R\) in L atm/(mol K) is approximately: \[R = 0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}\] Plugging in the known values to the ideal gas law equation, we can now solve for the moles of air (\(n\)): \[n = \frac{P V}{R T} = \frac{0.9869 \mathrm{atm} \cdot 1.0 \mathrm{~L}}{0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}} \cdot 298.15\mathrm{K}} = 0.0402 \mathrm{mol}\]

Now that we know the total number of moles of air in the given volume, we can use the fraction of CO in the air to determine the number of moles of CO present in 1.0 L of air: \[0.0402\ \mathrm{mol} \cdot \frac{3.0}{1,000,000} = 1.21 \times 10^{-4}\ \mathrm{mol}\]

Finally, we can multiply the number of moles of CO by Avogadro's number to obtain the number of CO molecules in 1.0 L of air: \[1.21 \times 10^{-4}\ \mathrm{mol} \cdot 6.022 \times 10^{23}\ \frac{\mathrm{molecules}}{\mathrm{mol}} = 7.27 \times 10^{19}\ \mathrm{molecules}\] In 1.0 L of this air at a pressure of 100 kPa and a temperature of 25°C, there are approximately \(7.27 \times 10^{19}\) CO molecules.