# Answer to Question #225926 in General Chemistry for William

Question #225926

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

1
2021-08-24T02:20:27-0400

Henry's law:

x = "\\dfrac{p}{H}x= \n\nH\n\np"

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 \n\n0"

=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 \n\n0"

= "3.64\u00d710 \n\n4\n\n \n\n0.21\n\n\u200b\t\n\n \u22485.8\u00d710 ^\n\n{\u22126}"

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 _\n\n0\n\n\u200b\t\n\n \u00d7555.6 moles=5.8\u00d710 \n\n\u22126\n\n \u00d7555.6 moles\u22483.22\u00d710 \n\n\u22123\n\n 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"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!