The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 10⁵ atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is:

According to Henry's law :

P = K_H x
where P is the partial pressure of the gas, K_H is Henry's law constant and x is the mole fraction of the gas in the solution.
Given:
K_H = 1.0 × 10⁵ atm
Mole fraction of N2 in air = 0.8
Pressure = 5 atm
Number of moles of water = 10 moles
Partial pressure of N2 = 0.8 × 5 atm = 4 atm
Therefore, x = P/K_H = 4 atm / (1.0 × 10⁵ atm) = 4 × 10⁻⁵
Let n be the number of moles of N2 dissolved in 10 moles of water.
Then, the mole fraction of N2 in the solution is given by:
x = n / (n + 10)
4 × 10⁻⁵ = n / (n + 10)
4 × 10⁻⁵ (n + 10) = n
4 × 10⁻⁵ n + 4 × 10⁻⁴ = n
n (1 - 4 × 10⁻⁵) = 4 × 10⁻⁴
n ≈ 4 × 10⁻⁴ moles
Therefore, the number of moles of N2 dissolved in 10 moles of water is approximately 4 × 10⁻⁴ moles.