Solutions Part 2

Concept:Henry's law states that the partial pressure of a gas above a liquid is directly proportional to its mole fraction in the liquid.Explanation:We are given that mole percentage of $\mathrm{N}_2$ in air is $80\%$ at 298 K.Total pressure of air over water is 10 atm.So the partial pressure of $\mathrm{N}_2$ is $P_{\mathrm{N}_2} = 0.80 \times 10 = 8$ atm.Convert this pressure to mmHg because $K_H$ is given in mmHg: $8 \text{ atm} \times 760 \text{ mmHg/atm} = 6080 \text{ mmHg}$.Henry's law constant $K_H = 6.5 \times 10^7 \text{ mmHg}$.Using Henry's law: $P_{\mathrm{N}_2} = K_H \cdot X_{\mathrm{N}_2}$.Thus, $X_{\mathrm{N}_2} = \frac{P_{\mathrm{N}_2}}{K_H} = \frac{6080}{6.5 \times 10^7}$.Calculate: $6080 \div 6.5 \approx 935.38$, then $X_{\mathrm{N}_2} = 935.38 \times 10^{-7} = 9.35 \times 10^{-5}$.Answer:The mole fraction of $\mathrm{N}_2$ in water is $9.35 \times 10^{-5}$, which corresponds to option A.