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He henry's law constant of oxygen in water at 25 °c is 773 atm mol-1kg h2o. calculate the molality of oxygen in water under a partial pressure of 0.20 at
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He henry's law constant of oxygen in water at 25 °c is 773 atm mol-1kg h2o. calculate the molality of oxygen in water under a partial pressure of 0.20 at

User Blm
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1 Answer

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Henry law:

P = KC
P = solvent vapour pressure
K = Henry constant
C = conc of solute

So, 0.2 = 773 C
so, C = 0.2 / 773 = 2.5 x 10^{-4} molal
User Piotr Malec
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