91Ó°ÊÓ

First, let's find the partial pressure of the \(\mathrm{CO}_{2}\) in the expired air at its peak. The mole fraction of the \(\mathrm{CO}_{2}\) in the expired air is 4.6% by volume, which can be converted to a ratio: Mole Fraction of \(\mathrm{CO}_{2} = \frac{4.6}{100} = 0.046\) Since the total pressure is given as 101.3 kPa, we can now find the partial pressure of \(\mathrm{CO}_{2}\) using the formula: Partial Pressure of \(\mathrm{CO}_{2} = 0.046 \times 101.3 \, \mathrm{kPa}\) Partial Pressure of \(\mathrm{CO}_{2} = 4.6598 \, \mathrm{kPa}\)

Now let's find the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak. To do this, we will first convert the given temperature of \(37^{\circ} \mathrm{C}\) to Kelvin: Temperature in \(K = 37^{\circ} \mathrm{C} + 273.15 \,K\) Temperature in \(K = 310.15 \,K\) Assuming that \(\mathrm{CO}_{2}\) behaves as an ideal gas, we can use the Ideal Gas Law to find the molarity given partial pressure and temperature as: \(PV = nRT\) Where \(P\) is the partial pressure, \(V\) is the volume, n is the amount of substance in moles, \(R\) is the ideal gas constant (8.314 J/molâ‹…K), and \(T\) is the temperature in Kelvin. We are interested in molarity, which is given by: Molarity \(= \frac{n}{V}\) Rearranging the Ideal Gas Law: \(\frac{n}{V} = \frac{P}{RT}\) By plugging the values of the partial pressure of \(\mathrm{CO}_{2}\) and the temperature in Kelvin, we get: Molarity \(= \frac{4.6598 \, \mathrm{kPa}}{(8.314 \times 10^{-3} \, \frac{\mathrm{kPa}\cdot\mathrm{L}}{\mathrm{mol}\cdot\mathrm{K}} )(310.15 \, K)}\) Molarity \(= 0.0182 \, \frac{\mathrm{mol}}{\mathrm{L}}\) So the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak is 0.0182 M.