At a certain temperature the Henry's law constant of a gas in water is $K_{H} = 8.5 \times 1 0^{4} bar$ (with $p = K_{H} x$ , $x$ the mole fraction of the gas in solution). Calculate the amount of gas, in millimoles, dissolved in $1.00 kg$ of water at a partial pressure of $4.0 bar$ . Take $M (H X_{2} O) = 18 g mo l^{- 1}$ . Give the answer to two decimal places.

Numerical answer type, answer in mmol, tolerance 0.01.

Answer: 2.61 mmol

x = p / K_{H} = 4.0/ (8.5 \times 1 0^{4}) = 4.706 \times 1 0^{- 5}

. Moles of water

= 1000/18 = 55.56

, and

n = \frac{x}{1 - x} \times 55.56

. This gives

2.615 mmol