Some soft drinks contain up to 85 ppm oxygen. (a) What is this concentration in mol/L? (b) What partial pressure of \(\mathrm{O}_{2}\) above water is needed to obtain \(85 \mathrm{ppm} \mathrm{O}_{2}\) in water at $10^{\circ} \mathrm{C} ?\( (The Henry's law constant for \)\mathrm{O}_{2}$ at this temperature is $\left.1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa} .\right)$

To find the concentration in mol/L, we need to convert the given concentration from ppm to mol/L. The given concentration is 85 ppm of oxygen, which means there are 85 grams of O2 in 1 million grams of water: Concentration in g/L = \(\frac{85 \mathrm{g}}{10^6 \mathrm{g}} * 10^3 \mathrm{g} = 0.085 \mathrm{g/L}\) Now, we will convert the concentration in g/L to mol/L. The molar mass of oxygen, O2, is \(2 * 16.0 = 32.0\) grams/mol. Concentration in mol/L = \(\frac{0.085 \mathrm{g/L}}{32.0 \mathrm{~g/mol}} = 2.66 \times 10^{-3} \mathrm{mol/L}\)

To find the partial pressure of O2, we will use Henry's law formula, which states that: \(P_\mathrm{O2} = K_\mathrm{H} * C\) Where: - \(P_\mathrm{O2}\): Partial pressure of O2 above water - \(K_\mathrm{H}\): Henry's law constant - C: Concentration of O2 in the water In this exercise, we are given the Henry's law constant for O2 at 10°C as \(1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\) To find the partial pressure of \(\mathrm{O}_{2}\) above water, we will plug the Henry's law constant value of O2 and the concentration of O2 in the water: \(P_\mathrm{O2} = \left(1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\right) * \left(2.66 \times 10^{-3} \mathrm{mol/L}\right)\) To match the units on both sides, we need to convert the concentration of O2 from mol/L to mol/\(\mathrm{m}^{3}\): \(2.66 \times 10^{-3} \mathrm{mol/L}\) = \(2.66 \times 10^{-3} \mathrm{mol/L} * \frac{10^3 \mathrm{L}}{1 \mathrm{m}^3} = 2.66 \mathrm{mol/m}^3\) Now, we can calculate the partial pressure of \(\mathrm{O}_{2}\): \(P_\mathrm{O2} = \left(1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\right) * \left(2.66 \mathrm{mol/m}^3\right) = 4.49 \times 10^{-5} \mathrm{Pa}\) The partial pressure of \(\mathrm{O}_{2}\) above water to obtain \(85 \mathrm{ppm}\) of \(\mathrm{O}_{2}\) in water at \(10^{\circ} \mathrm{C}\) is \(4.49 \times 10^{-5} \mathrm{Pa}\).