Air (taken as 21% O2 by mole) at a total pressure of 5.0 bar is in equilibrium with water at a temperature where K_H(O2) = 4.5x10^4 bar (law p = K_H x). Treat water as 1000 g with M = 18 g mol^-1 and M(O2) = 32 g mol^-1. Find the mass of O2 dissolved, in mg, to one decimal place.

JEE Advanced levelNumerical answerDifficulty 3 of 3so012

Air (taken as $21%$ $O X_{2}$ by mole) at a total pressure of $5.0 bar$ is in equilibrium with water at a temperature where $K_{H} (O X_{2}) = 4.5 \times 1 0^{4} bar$ (law $p = K_{H} x$ ). Treat water as $1000 g$ with $M = 18 g mo l^{- 1}$ and $M (O X_{2}) = 32 g mo l^{- 1}$ . Find the mass of $O X_{2}$ dissolved, in mg, to one decimal place.

Numerical answer type, answer in mg, tolerance 0.1.

p_{O X_{2}} = 0.21 \times 5.0 = 1.05 bar

, so

x = 1.05/ (4.5 \times 1 0^{4}) = 2.333 \times 1 0^{- 5}

. Moles of water

= 55.56

, so

n_{O X_{2}} = \frac{x}{1 - x} \times 55.56 = 1.296 \times 1 0^{- 3} mol

. Mass

= n \times 32 \times 1000 = 41.48 mg