At 298 K, the mole percentage of $$N_{2}$$(g) in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $$N_{2}$$(g) in water at 298 K?
($$K_{H}$$ for $$N_{2}$$ is $$6.5 \times 10^{7}$$ mm Hg)

We need to find the mole fraction of N₂ dissolved in water using Henry's law.

Mole percentage of N₂ in air = 80%, Pressure = 10 atm

$$K_H$$ for N₂ = $$6.5 \times 10^7$$ mm Hg

Find partial pressure of N₂.

$$P_{N_2} = 0.80 \times 10 = 8$$ atm = $$8 \times 760 = 6080$$ mm Hg

Apply Henry's law.

$$P_{N_2} = K_H \times x_{N_2}$$

$$x_{N_2} = \frac{P_{N_2}}{K_H} = \frac{6080}{6.5 \times 10^7} = 9.35 \times 10^{-5}$$

Therefore, the mole fraction is Option 3: $$9.35 \times 10^{-5}$$.