Breathing air that contains 4.0\(\%\) by volume \(\mathrm{CO}_{2}\) over time causes rapid breathing, throbbing headache, and nausea, among other symptoms. What is the concentration of \(\mathrm{CO}_{2}\) in such air in terms of (a) mol percentage, (b) molarity, assuming 1 atm pressure and a body temperature of \(37^{\circ} \mathrm{C} ?\)

The given volume percentage of CO2 is 4.0%. In a mixture of gases, the volume percentage is equal to the mol percentage. Hence, mol percentage of CO2 is 4.0%.

We will use the Ideal Gas Law formula: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Given: - Pressure (P) = 1 atm - Temperature (T) = 37°C = 310.15 K (converting to Kelvin) - Volume percentage of CO2 = 4.0% First, we need to find the partial pressure of CO2 in the air using the given volume percentage of CO2. Partial pressure of CO2 = total pressure × volume percentage of CO2: \[P_{\mathrm{CO}_2} = 1\,\mathrm{atm} \times 0.04\] \[P_{\mathrm{CO}_2} = 0.04\,\mathrm{atm}\] Now, we will use the Ideal Gas Law formula to find the molarity of CO2. Rearranging the formula for molarity (n/V): \[\frac{n}{V} = \frac{P}{RT}\] Substitute the values: \[\frac{n}{V} = \frac{0.04\,\mathrm{atm}}{(0.0821\,\mathrm{L\,atm/(mol\,K)})\times 310.15\,\mathrm{K}}\] Now, calculate the molarity: \[\frac{n}{V} = 0.0016\,\mathrm{mol/L}\] So, the concentration of CO2 in terms of (a) mol percentage is 4.0%, and (b) molarity is 0.0016 mol/L.