Dry air contains 78% N2 by mole. Air saturated with water vapour (vapour pressure 0.040 atm at the temperature) is in equilibrium with water at a total pressure of 1.00 atm. Take K_H(N2)=8.5x10^4 atm (for p=K_Hx), treat the solution as dilute so that n_N2approx xx(1000/18) per kg of water, and use 28 g mol^-1 for N2. Find the mass of dissolved N2 in mg per kg of water, to one decimal place.

JEE Advanced levelNumerical answerDifficulty 3 of 3so146

Dry air contains $78%$ $N X_{2}$ by mole. Air saturated with water vapour (vapour pressure $0.040$ atm at the temperature) is in equilibrium with water at a total pressure of $1.00$ atm. Take $K_{H} (N X_{2}) = 8.5 \times 1 0^{4}$ atm (for $p = K_{H} x$ ), treat the solution as dilute so that $n_{N X_{2}} \approx x \times (1000/18)$ per kg of water, and use $28 g mo l^{- 1}$ for $N X_{2}$ . Find the mass of dissolved $N X_{2}$ in mg per kg of water, to one decimal place.

Numerical answer type, answer in mg/kg, tolerance 0.1.

Moist air has

N X_{2}

partial pressure

0.78 \times (1.00 - 0.040) = 0.7488

atm. Then

x = 0.7488/ (8.5 \times 1 0^{4}) = 8.809 \times 1 0^{- 6}

. Moles per kg water

= 8.809 \times 1 0^{- 6} \times 55.56 = 4.894 \times 1 0^{- 4}

mol, mass

= 4.894 \times 1 0^{- 4} \times 28 \times 1000 = 13.7

mg.