Answer to Question #225926 in General Chemistry for William

Question

How can I calculate the quantity of oxygen in freshwater in a 50L bucket at an altitude of 5000m above sea level

Answer 1

Henry's law:

x = $\dfrac{p}{H}x= H p$

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where xx is the mole fraction of the oxigen in the water, pp is the partial oxygen preassure in air (in atm) and H = 3.64\times 10^{4}H=3.64×10

4

is Henry's constant for oxygen at 15°c.

1. The partial oxygen preassure in air at sea level is:

$p_0 = 1\space atm\times 0.21 = 0.21\space atmp 0$

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=1 atm×0.21=0.21 atm

where 0.210.21 represents the fraction of oxygen in air.

Thus:

$x_0 = \dfrac{0.21}{3.64\times 10^{4}} \approx 5.8\times 10^{-6}x 0$

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= $3.64×10 4 0.21 ​ ≈5.8×10 ^ {−6}$

One liter of water contains 55.56 moles of water. Thus, 10 liters will contain 555.6 moles. Thus, the quantity of oxygen will be.

$n_0 = x_0\times 555.6 \space moles = 5.8\times 10^{-6}\times 555.6\space moles \approx 3.22\times 10^{-3}\space molesn _0$

$=x _ 0 ​ ×555.6 moles=5.8×10 −6 ×555.6 moles≈3.22×10 −3 moles$

The partial oxygen preassure in air at 5000 m.a.s is:

$$p_0 = 0.53\space atm\times 0.21 = 0.11\space atm$$

where $0.21$ represents the fraction of oxygen in air.

Thus:

$$x_0 = \dfrac{0.11}{3.64\times 10^{4}} \approx 3.02\times 10^{-6}$$

One liter of water contains 55.56 moles of water. Thus, 10 liters will contain 555.6 moles. Thus, the quantity of oxygen will be:

$$n_0 = x_0\times 555.6 \space moles = 3.02\times 10^{-6}\times 555.6\space moles \approx 1.68\times 10^{-3}\space moles$$