A bottle holds 330 g of water saturated with CO2 at 4.0 bar (sealed). For CO2 in water, K_H = 1670 bar (law p = K_H x). After opening, the water equilibrates with air containing CO2 at a partial pressure of 4.0x10^-4 bar. Take M(H2O) = 18 g mol^-1 and use x/(1-x)=n_CO2/n_H2O without further approximation. Find the amount of CO2 released, in millimoles, to the nearest integer.

JEE Advanced levelNumerical answerDifficulty 3 of 3so014

A bottle holds $330 g$ of water saturated with $CO X_{2}$ at $4.0 bar$ (sealed). For $CO X_{2}$ in water, $K_{H} = 1670 bar$ (law $p = K_{H} x$ ). After opening, the water equilibrates with air containing $CO X_{2}$ at a partial pressure of $4.0 \times 1 0^{- 4} bar$ . Take $M (H X_{2} O) = 18 g mo l^{- 1}$ and use $x / (1 - x) = n_{CO X_{2}} / n_{H X_{2} O}$ without further approximation. Find the amount of $CO X_{2}$ released, in millimoles, to the nearest integer.

Numerical answer type, answer in mmol, tolerance 0.5.

n_{H X_{2} O} = 330/18 = 18.33 mol

. Sealed:

x_{1} = 4.0/1670 = 2.3952 \times 1 0^{- 3}

and

n_{1} = x_{1} / (1 - x_{1}) \times 18.33 = 0.04402 mol

. Opened:

x_{2} = 4.0 \times 1 0^{- 4} /1670

gives

n_{2} = 4.39 \times 1 0^{- 6} mol

. Released

= (n_{1} - n_{2}) \times 1000 = 44.01 mmol

, i.e. about 44 mmol.