At 298 K, the Henry\'s law constant for oxygen is 0.00130 M/atm. Air is 21.0% ox

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At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen. At 298K, what is the solubility of oxygen in water exposed to air at 1.00 atm? At 298K, what i the solubility of oxygen in water exposed to air at 0.891 atm? leaming.com/ibisoms/mod/ibis/view.php?id 3165672 2/7/2017 104s AM 48 MM100 03 Gradebook Print Caloaator Periodic Table Question 6 of 24 Sapling Learning AL298 K. what is the solubility of oxygen in water exposed to air at 1.00 atm? At 298 K. what is the solubility of oxygen in water exposed to air at 0891 atm? If atmospheric pressure suddenly changes from 1.00 atm to 0.891 atrm at 298 K. how much oxygen wil be from 2.50 L of water in an unsealed container? mol O Previous ® Give up s View Solution Check Answer 0 Next HExt about us careers partners Privacy policy terms of use contact us he 0

Explanation / Answer

Given that the Air is 21.0% oxygen.

Hence, the partial pressure of oxygen in the atmosphere = 1 atm x 21% O2 = 0.21 atm

a. At 298 K and 1 atm, the solubility of oxygen in the water = 0.00130 M/atm x 0.21 x 1 atm = 0.000273 mole / L

b. At 298 K and 0.891 atm, the solubility of oxygen in the water = 0.00130 M/atm x 0.21 x 0.891 atm = 0.000243 mole / L

c. If atmospheric pressure suddenly changes from 1.00 atm to 0.891 atm at 298 K, the change in solubility is 0.000273 M - 0.000243 M = 0.00003 mole / L

Further, Concentration = mole / L,

                            mole = concentration x volume

                                   = 0.00003 mole / L x 2.5 L = 0.000075 mole

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