Exhaled air contains \(74.5 \% \mathrm{~N}_{2}, 15.7 \% \mathrm{O}_{2}, 3.6 \% \mathrm{CO}_{2}\), and \(6.2 \% \mathrm{H}_{2} \mathrm{O}\) (mole percent). (a) Calculate the molar mass of exhaled air. (b) Calculate the density of exhaled air at \(37^{\circ} \mathrm{C}\) and \(757 \mathrm{~mm} \mathrm{Hg}\), and compare the value obtained with that for ordinary air (MM = \(29.0 \mathrm{~g} / \mathrm{mol}\) ).

We need to know the molar masses (in g/mol) of each component present in the exhaled air. They are as follows: - \(\mathrm{N}_{2}: 28.02 ~\mathrm{g/mol}\) - \(\mathrm{O}_{2}: 32.00 ~\mathrm{g/mol}\) - \(\mathrm{CO}_{2}: 44.01 ~\mathrm{g/mol}\) - \(\mathrm{H}_{2}\mathrm{O}: 18.02 ~\mathrm{g/mol}\)

We will now use the mole percentage of each component to find the molar mass of exhaled air. The calculation is as follows: Molar mass of exhaled air = \((74.5 \% \times 28.02) + (15.7 \% \times 32.00) + (3.6 \% \times 44.01) + (6.2 \% \times 18.02)\) Molar mass of exhaled air = \((0.745 \times 28.02) + (0.157 \times 32.00) + (0.036 \times 44.01) + (0.062 \times 18.02)\) Molar mass of exhaled air = \(29.09 ~\mathrm{g/mol}\)

We will apply the ideal gas law represented as \(PV=nRT\), where: - P is the pressure in atm - V is the volume in liters - n is the number of moles - R is the ideal gas constant (\(0.0821 ~\mathrm{L \cdot atm} / (\mathrm{K} \cdot\mathrm{mol})\)) - T is the temperature in Kelvin First, we need to convert the temperature to Kelvin and the pressure to atm: - Temperature: \((37 + 273.15) = 310.15 ~\mathrm{K}\) - Pressure: \(757~\mathrm{mmHg} \times \dfrac{1~\mathrm{atm}}{760 ~\mathrm{mmHg}} = 0.996 ~\mathrm{atm}\) Now, let's find the density. The density of a gas is given by the formula \(\rho = \dfrac{n \cdot MM}{V}\). Replacing n from the ideal gas law and modifying, we get: \(\rho = \dfrac{P \cdot MM}{R \cdot T}\) Now, we can plug the values and calculate the density of exhaled air: \(\rho = \dfrac{0.996 ~\mathrm{atm} \times 29.09 ~\mathrm{g/mol}}{0.0821 ~\mathrm{L \cdot atm / (K \cdot mol)} \times 310.15 ~\mathrm{K}} = 1.146 ~\mathrm{g/L}\)

We have calculated the density of exhaled air and now we will compare it with the density of ordinary air (MM = \(29.0 ~\mathrm{g/mol}\)). Using the same formula as in Step 3, we can find the density of ordinary air: \(\rho_{ordinary} = \dfrac{0.996 ~\mathrm{atm} \times 29.0 ~\mathrm{g/mol}}{0.0821 ~\mathrm{L \cdot atm / (K \cdot mol)} \times 310.15 ~\mathrm{K}} = 1.142 ~\mathrm{g/L}\) Thus, the density of exhaled air (\(1.146 ~\mathrm{g/L}\)) is slightly higher than the density of ordinary air (\(1.142 ~\mathrm{g/L}\)).